UNIVERSITY  OF  CALIFORNIA 
AT  LOS  ANGELES 


Gift     Of 

Prof .Harry  Thomas  Cory 


-rA 


PART    III. 


STRENGTH  OF  MATERIALS. 

[OB  MECHANICS  OF  MATERIALS]. 
CHAPTER  I. 

ELEMENTARY  STRESSES  AND  STRAINS. 

178.  Deformation  of  Solid  Bodies. — In  the  preceding  por- 
tions of  this  work,  what  was  called  technically  a  "  rigid 
body,"  was  supposed  incapable  of  changing  its  form,  i.e., 
the  positions  of  its  particles  relatively  to  each  other,  under 
the  action  of  any  forces  to  be  brought  upon  it.  This  sup- 
position was  made  because  the  change  of  form  which  must 
actually  occur  does  not  appreciably  alter  the  distances, 
angles,  etc.,  measured  in  any  one  body,  among  most  of 
the  pieces  of  a  properly  designed  structure  or  machine. 
To  show  how  the  individual  pieces  of  such  constructions 
should  be  designed  to  avoid  undesirable  deformation  or 
injury  is  the  object  of  this  division  of  Mechanics  of  En- 
gineering, viz.,  the  Strength  of  Materials. 

^Q       ^L  g^ 

X^X —      — 5 —  — ^^/ 

Fm.  198.    §  17& 

As  perhaps  tne  simplest  instance  of  the  deformation  or 
distortion  of  a  solid,  let  us  consider  the  case  of  a  prismatic 
rod  in  a  state  of  tension,  Fig.  192  (link  of  a  surveyor's 


1  i 

I   i^a 


196  MECHANICS  OF   ENGINEERING. 

chain,  e.g.).  The  pull  at  each  end  is  P,  and  the  body  is 
said  to  be  under  a  tension  of  P  (Ibs.,  tons,  or  other  unit), 
not  2P.  Let  ABGD  be  the  end  view  of  an  elementary 
parallelepiped,  originally  of  square  section  and  with  faces 
at  45°  with  the  axis  of  the  prism.  It  is  now  deformed,  the 
four  faces  perpendicular  to  the  paper  being  longer*  than 
before,  while  the  angles  BAD  and  BCD,  originally  right 
angles,  are  now  smaller  by  a  certain  amount  d,  ABG  and 
ADC  larger  by  an  equal  amount  3.  The  element  is  said 
to  be  in  a  state  of  strain,  viz.:  the  elongation  of  its  edges 
(parallel  to  paper)  is  called  a  tensile  strain,  while  the  alter- 
ation in  the  angles  between  its  faces  is  called  a  shearing 
strain,  or  angular  distortion  (sometimes  also  called  a  slid- 
ing, or  tangential,  strain,  since  BC  has  been  made  to  slide, 
relatively  to  AD,  and  thereby  caused  the  change  of  angle). 
[This  use  of  the  word  strain,  to  signify  change  of  form  and 
not  the  force  producing  it,  is  of  recent  adoption  among 
many,  though  not  all,  technical  writers.] 

179.  Strains.  Two  Kinds  Only. — Just  as  a  curved  line  may 
be  considered  to  be  made  up  of  small  straight-line  ele- 
ments, so  the  substance  of  any  solid  body  may  be  consid- 
ered to  be  made  up  of  small  contiguous  parallelepipeds, 
whose  angles  are  each  90°  before  the  body  is  subjected  to 
the  action  of  forces,  but  which  are  not  necessarily  cubes. 
A  line  of  such  elements  forming  an  elementary  prism  is 
sometimes  called  a,  fibre,  but  this  does  not  necessarily  imply 
a  fibrous  nature  in  the  material  in  question.  The  system 
of  imaginary  cutting  surfaces  by  which  the  body  is  thus 
subdivided  need  not  consist  entirely  of  planes  ;  in  the  sub- 
ject of  Torsion,  for  instance,  the  parallelopipedical  ele- 
ments considered  lie  in  concentric  cylindrical  shells,  cut 
both  by  transverse  and  radial  planes. 

Since  these  elements  are  taken  so  small  that  the  only 
possible  changes  of  form  in  any  one  of  them,  as  induced 
by  a  system  of  external  forces  acting  on  the  body,  are 

*  When  a  is  nearly  0°  (or  90")  BG  and  AD  (or  AB  and  DC)  are  shorter 
than  before,  on  account  of  lateral  contraction;  see  §  193. 


ELEMENTARY   STRESSES,    ETC.  197 

elongations  or  contractions  of  its  edges,  and  alteration  of 
its  angles,  there  are  but  two  kinds  of  strain,  elongation 
(contraction,  if  negative)  and  shearing. 

180.  Distrihuted  Forces  or  Stresses.  —  In  the  matter  preced- 
ing this  chapter  it  has  sufficed  for  practical  purposes  to 
consider  a  force  as  applied  at  a  point  of  a  body,  but  in 
reality  it  must  be  distributed  over  a  definite  area  ;  for 
otherwise  the  material  would  be  subjected  to  an  infinite 
force  per  unit  of  area.  (Forces  like  gravity,  magnetic  at- 
traction, etc.,  we  have  already  treated  as  distributed  over 
the  mass  of  a  body,  but  reference  is  now  had  particularly 
to  the  pressure  of  one  body  against  another,  or  the  action 
of  one  portion  of  the  body  on  the  remainder.)  For  in- 
stance, sufficient  surface  must  be  provided  between  the 
end  of  a  loaded  beam  and  the  pier  on  which  it  rests  to 
avoid  injury  to  either.  Again,  too  small  a  wjre  must  not 
be  used  to  sustain  a  given  load,  or  the  tension  per  unit 
of  area  of  its  cross  section  becomes  sufficient  to  rupture 
it. 

Stress  is  distributed  force,  and  its  intensity  at  any  point 
of  the  area  is 


where  dF  is  a  small  area  containing  the  point  and  dP  the 
force  coming  upon  that  area.  If  equal  dP's  (all  parallel) 
act  on  equal  dfsof  a  plane  surface,  the  stress  is  said  to 
be  of  uniform  intensity,  which  is  then 


where  P=  total  force  and  F  the  total  area  over  which  it 
acts.  The  steam  pressure  on  a  piston  is  an  example  of 
stress  of  uniform  intensity. 


198  MECHANICS    OF   ENGINEERING. 

For  example,  if  a  force  P=  28800  Ibs,  is  uniformly  dis- 
tributed over  a  plane  area  of  F=  72  sq.  inches,  or  ^  of  a 
sq.  foot,  the  intensity  of  the  stress  is 

=400  Ibs.  per  sq.  inch, 


72 

(or  p  =  28800-^=57600  Ibs.  per  sq.  foot,  or  _p=14400-5» 
^=28.8  tons  per  sq.  ft.,  etc.). 

181.  Stresses  on  an  Element  ;  of  Two  Kinds  Only.  —  When  a 
solid  body  of  any  material  is  in  equilibrium  under  a  sys- 
tem of  forces  which  do  not  rupture  it,  not  only  is  its  shape 
altered  (i.e.  its  elements  are  strained],  and  stresses  pro- 
duced on  those  planes  on  which  the  forces  act,  but  other 
stresses  also  are  induced  on  some  or  all  internal  surfaces 
which  separate  element  from  element,  (over  and  above  the 
forces  with  which  the  elements  may  have  acted  on  each 
other  before  the  application  of  the  external  stresses  or 
"  applied  forces  ").  So  long  as  the  whole  solid  is  the  "free 
body  "  under  consideration,  these  internal  stresses,  being 
the  forces  with  which  the  portion  on  one  side  of  an  imag- 
inary cutting  plane  acts  on  the  portion  on  the  other  side, 
do  not  appear  in  any  equation  of  equilibrium  (for  if  intro- 
duced they  would  cancel  out);  but  if  we  consider  free  a 
portion  only,  some  or  all  of  whose  bounding  surfaces  are 
cutting  planes  of  the  original  body,  the  stresses  existing 
on  these  planes  are  brought  into  the  equations  of  equilib- 
rium. 

Similarly,  if  a  single  element  of  the  body  is  treated  by 
itself,  the  stresses  on  all  six  of  its  faces,  together  with  its 
weight,  form  a  balanced  system  of  forces,  the  body  being 
supposed  at  rest. 


Fie.  193. 


ELEMENTARY   STRESSES,    ETC. 


199 


As  an  example  of  internal  stress,  consider  again  the  case 
of  a  rod  in  tension  ;  Fig.  193  shows  the  whole  rod  (or  eye- 
bar)  free,  the  forces  P  being  the  pressures  of  the  pins  in 
the  eyes,  and  causing  external  stress  (compression  here) 
on  the  surfaces  of  contact.  Conceive  a  right  section  made 
through  US,  far  enough  from  the  eye,  G,  that  we  may  con- 
sider the  internal  stress  to  be  uniform  in  this  section,  and 
consider  the  portion  BSG  as  a  free  body,  in  Fig.  194.  The 
stresses  on  US,  now  one  of  the  bounding  surfaces  of  the 
free  body,  must  be  parallel  to  P,  i.e.,  normal  to  US', 
(otherwise  they  would  have  components  perpendicular  to 
P,  which  is  precluded  by  the  necessity  of  2Y  being  =  0, 
and  the  supposition  of  uniformity.)  Let  F  =  the  sec- 


Fio.  195. 


tional  area  RS,  and  p  =  the  stress  per  unit  of  area ;  then 


-T.Y=  0  gives  P 


Fp,  i.e.,  p=  £ 
F 


•(2) 


The  state  of  internal  stress,  then,  is  such  that  on  planes 
perpendicular  to  the  axis  of  the  bar  the  stress  is  tensile  and 
normal  (to  those  planes).  Since  if  a  section  were  made 
oblique  to  the  axis  of  the  bar,  the  stress  would  still  be 
parallel  to  the  axis  for  reasons  as  above,  it  is  evident  that 
on  an  oblique  section,  the  stress  has  components  both  nor- 
mal and  tangential  to  the  section,  the  normal  component 
being  a  tension. 


200 


MECHANICS   OF  ENGINEERING. 


The  presence  of  the  tangential  or  shearing  stress  in  ob- 
lique sections  is  rendered  evident  by  considering  that  if  an 
oblique  dove-tail  joint  were  cut  in  the  rod,  Fig.  195,  the 
shearing  stress  on  its  surfaces  may  be  sufficient  to  over- 
come friction  and  cause  sliding  along  the  oblique  plane. 

If  a  short  prismatic  block  is  under  the  compressive  ac- 
tion of  two  forces,  each  =P  and  applied  centrally  in  one 
base,  we  may  show  that  the  state  of  internal  stress  is  the 
same  as  that  of  the  rod  under  tension,  except  that  the  nor- 
mal stresses  are  of  contrary  sign,  i.e.,  compressive  instead 
of  tensile,  and  that  the  shearing  stresses  (or  tendency  to 
slide)  on  oblique  planes  are  opposite  in  direction  to  those 
in  the  rod. 

Since  the  resultant  stress  on  a  given  internal  plane  of  a 
body  is  fully  represented  by  its  normal  and  tangential 
components,  we  are  therefore  justified  in  considering  but 
two  kinds  of  internal  stress,  normal  or  direct,  and  tangen- 
tial or  shearing. 

182.  Stress  on  Oblique  Section  of  Rod  in  Tension. — Consider 
free  a  small  cubic  element  whose 
edge  =a  in  length;  it  has  two 
faces  parallel  to  the  paper,  being 
taken  near  the  middle  of  the  rod 
in  Fig.  192.  Let  the  angle  which 
the  face  AS,  Fig.  196,  makes  with 
the  axis  of  the  rod  be  =  a.  This 
angle,  for  our  present  purpose,  is 
considered  to  remain  the  same 
while  the  two  forces  P  are  acting, 
as  before  their  action.  The  re- 
sultant stress  on  the  face  AB  hav- 
ing an  intensity  p^P-^-F,  (see  eq. 
2)  per  unit  of  transverse  section 
of  rod,  is  =  p  (a  sin  a)  a.  Hence 
FlG-  197-  its  component  normal  to  AB  is 

pa2  sin2  a  ;  and  the  tangential  or  shearing  component  along 


ELEMENTARY    STRESSES,    ETC.  201 

A B  =-jt>a2  sin  a  cos  a.      Dividing  by  the  area,  a2,  we  have 
the  following : 

For  a  rod  in  simple  tension  we  have,  on  a  plane  making 
an  angle,  a,  with  the  axis : 

a  Normal  Stress  =p  sin2  a  per  unit  of  area      .         .     (1) 
and  a  Shearing  Stress  =p  sin  a  cos  a  per  unit  of  area    .     (2) 

"  Unit  of  area  "  here  refers  to  the  oblique  plane  in  ques- 
tion, while  p  denotes  the  normal  stress  per  unit  of  area  of 
a  transverse  section,  i.e.,  when  a=90°,  Fig.  194. 

The  stresses  on  CD  are  the  same  in  value,  as  on  AB, 
while  for  BG  and  AD  we  substitute  90°— a  for  a.  Fig. 
197  shows  these  normal  and  shearing  stresses,  and  also, 
much  exaggerated,  the  strains  or  change  of  form  of  the 
element  (see  Fig.  192). 

182a.  Relation  between  Stress  and  Strain. — Experiment 
shows  that  so  long  as  the  stresses  are  of  such  moderate 
value  that  the  piece  recovers  its  original  form  completely 
when  the  external  forces  which  induce  the  stresses  are  re- 
moved, the  following  is  true  and  is  known  as  Hookas  Law 
(stress  proportional  to  strain).  As  the  forces  P  in  Fig. 
193  (rod  in  tension)  are  gradually  increased,  the  elonga- 
tion, or  additional  length,  of  RK  increases  in  the  same 
ratio  as  the  normal  stress,  p,  on  the  sections  US  and  KN> 
per  unit  of  area  [§  191]. 

As  for  the  distorting  effect  of  shearing  stresses,  consider 
in  Fig.  197  that  since 

p  sin  a  cos  a  —  p  cos  (90° — a)  sin  (90° — a) 

the  shearing  stress  per  unit  of  area  is  of  equal  value  en  alt 
four  of  the  faces  (perpendicular  to  paper)  in  the  elementary 
block,  and  is  evidently  accountable  for  the  shearing  strain, 
i.e.,  for  the  angular  distortion,  or  difference,  d,  between 
90°  and  the  present  value  of  each  of  the  four  angles.  Ac- 
cording to  Hooke's  Law  then,  as  P  increases  within  tta 
limit  mentioned  above,  3  varies  proportionally  to 

p  sin  a  cos  a,  i.e.  to  the  stress. 


202  MECHANICS   OF  ENGINEERING. 

182b.  Example. — Supposing  the  rod  in  question  were  of 
a  kind  of  wood  in  which  a  shearing  stress  of  200  Ibs.  per 
sq.  inch  along  the  grain,  or  a  normal  stress  of  400  Ibs.  per 
sq.  inch,  perpendicular  to  a  fibre-plane  will  produce  rup- 
ture, required  the  value  of  «  the  angle  which  the  grain 
must  make  with  the  axis  that,  as  P  increases,  the  danger 
of  rupture  from  each  source  may  be  the  same.  This  re- 
quires that  200:400::^)  sin  a  cos  a  :p  sin2a,  i.e.  tan.  a  must 
=  2.000.'.«=63^°.  If  the  cross  section  of  the  rod  is  2  sq. 
inches,  the  force  P  at  each  end  necessary  to  produce  rup- 
ture of  either  kind,  when  a=63^°,  is  found  by  putting 
p  sin  a  cos  a=  20Q.'.p  =500.0  Ibs.  per  sq.  inch.  Whence,  since 
p=P+F,  P=1000  Ibs.  (Units,  inch  and  pound.) 

183.  Elasticity  is  the  name  given  to  the  property  which 
most  materials  have,  to  a  certain  extent,  of  regaining  their 
original  form  when  the  external  forces  are  removed.  If 
the  state  of  stress  exceeds  a  certain  stage,  called  the  Elastic 
Limit,  the  recovery  of  original  form  on  the  part  of  the  ele- 
ments is  only  partial,  the  permanent  deformation  being 
called  the  Set. 

Although  theoretically  the  elastic  limit  is  a  perfectly  defi- 
nite stage  of  stress,  experimentally  it  is  somewhat  indefi- 
nite, and  is  generally  considered  to  be  reached  when  the 
permanent  set  becomes  well  marked  as  the  stresses  are  in- 
creased and  the  test  piece  is  given  ample  time  for  recovery 
in  the  intervals  of  rest. 

The  Safe  Limit  of  stress,  taken  well  within  the  elastic 
limit,  determines  the  working  strength  or  safe  load  of  the 
piece  under  consideration.  E.g.,  the  tables  of  safe  loads 
of  the  rolled  wrought  iron  beams,  for  floors,  of  the  New 
Jersey  Steel  and  Iron  Co.,  at  Trenton,  are  computed  on 
the  theory  that  the  greatest  normal  stress  (tension  or  com- 
pression) occurring  on  any  internal  plane  shall  not  exceed 
12,000  Ibs.  per  sq.  inch  ;  nor  the  greatest  shearing  stress 
4,000  Ibs.  per  sq.  inch. 


ELEMENTARY   STRESSES,    ETC.  203 

The  intimate  Limit  is  reached  when  rupture  occurs. 

184.  The  Modulus  of  Elasticity  (sometimes  called  co-efficient 
of  elasticity)  is  the  number  obtained  by  dividing  the  stress 
per  unit  of  area  by  the  corresponding  relative  strain. 

Thus,  a  rod  of  wrought  iron  y2  sq.  inch  sectional  area 
being  subjected  to  a  tension  of  2^  tons  =5,000  Ibs.,  it  is 
found  that  a  length  which  was  six  feet  before  tension  is 
-=  6.002  ft.  during  tension.  The  relative  longitudinal  strain 
or  elongation  is  then=  (0.002)-=- 6=1 : 3,000  and  the  corres- 
ponding stress  (being  the  normal  stress  on  a  transverse 
plane)  has  an  intensity  of 

pt=P+F=  5,000-r-  >£  =10,000  Ibs.,  per  sq.  inch. 

Hence  by  definition  the  modulus  of  elasticity  is  (for  ten- 
sion) 

#t=jVf-£=10,000-r  3-^=30,000,000  Ibs.  per  sq.  inch,  (the 

sub-script  "  t "  refers  to  tension). 

It  will  be  noticed  that  since  E  is  an  abstract  number,  Et 
is  of  the  same  quality  as  pt,  i.e.,  Ibs.  per  sq.  inch,  or  one  di- 
mension of  force  divided  by  two  dimensions  of  length. 
(In  the  subject  of  strength  of  materials  the  inch  is  the 
most  convenient  English  linear  unit,  when  the  pound  is 
the  unit  of  force  ;  sometimes  the  foot  and  ton  are  used  to- 
gether.) 

The  foregoing  would  be  called  the  modulus  of  elasticity 
of  wrought  iron  in  tension  in  the  direction  of  the  fibre,  as 
given  by  the  experiment  quoted.  But  by  Hooke's  Law^> 
and  e  vary  together,  for  a  given  direction  in  a  given  ma- 
terial, hence  within  the  elastic  limit  E  is  constant  for  a  given 
direction  in  a  given  material.  Experiment  confirms  this 
appr  oxim  ately . 

Similarly,  the  modulus  of  elasticity  for  compression  E* 


204  MECHANICS   OF   ENGINEERING. 

in  a  given  direction  in  a  given  material  may  be  determined 
by  experiments  on  short  blocks,  or  on  rods  confined  lat- 
erally to  prevent  flexure. 

As  to  the  modulus  of  elasticity  for  shearing,  E*,  we 
divide  the  shearing  stress  per  unit  of  area  in  the  given 
direction  by  d  (in  it  measure)  the  corresponding  angular 
strain  or  distortion ;  e.g.,  for  an  angular  distortion  of  1°  or 
d=.0174,  and  a  shearing  stress  of  1,566  Ibs.  per  sq.  inch, 
we  have  EA=  gg=  9,000,000  Ibs.  per  sq.  inch. 

Unless  otherwise  specified,  by  modulus  of  elasticity  will 
be  meant  a  value  derived  from  experiments  conducted 
within  the  elastic  limit,  and  this,  whether  for  normal  stress 
or  for  shearing,  is  approximately  constant  for  a  given  di- 
rection in  a  given  substance.* 

185.  Isotropes. — This  name  is  given  to  materials  which 
are  homogenous  as  regards  their  elastic  properties.     In 
such  a  material  the  moduli  of  elasticity  are  individually 
the  same  for  all  directions.     E.g.,  a  rod  of  rubber  cut  out 
of  a  large  mass  will  exhibit  the  same  elastic  behavior  when 
subjected  to  tension,  whatever  its  original  position  in  the 
mass.     Fibrous  materials  like  wood  and  wrought  iron  are 
not  isotropic  ;  the  direction  of  grain  in  the  former   must 
always  be  considered.     The  "  piling  "  and  welding  of  nu- 
merous small  pieces  of  iron  prevent  the  resultant  forging 
from  being  isotropic. 

186.  Resilience  refers  to  the  potential  energy  stored  in  a 
body  held  under  external  forces  in  a  state  of  stress  which 
does  not  pass  the  elastic  limit.     On  its  release  from  con- 
straint, by  virtue  of  its  elasticity  it  can  perform  a  certain 
amount  of  work  called  the  resilience,  depending  in  amount 
upon  the  circumstances  of  each  case  and  the  nature  of  the 
material.     See  §  148. 

187.  General  Properties  of  Materials. — In  view  of  some  defi- 
nitions already  made  we  may  say  that  a  material  is  ductile 

*  The  moduli,  or  "  co-efficients,"  of  elasticity  as  used  by  physicists  are  well  explained 
in  Stewart  and  Gee's  Practical  Physics,  Vol.  I.,  pp.  164,  etc.  Their  "co-efficient  of 
rigidity"  is  our  Et. 


ELEMENTARY   STRESSES,    ETC.  205 

when  the  ultimate  limit  is  far  removed  from  the  elastic 
limit ;  that  it  is  brittle  like  glass  and  cast  iron,  when  those 
limits  are  near  together.  A  small  modulus  of  elasticity 
means  that  a  comparatively  small  force  is  necessary  to 
produce  a  given  change  of  form,  and  vice  versa,  but  implies 
little  or  nothing  concerning  the  stress  or  strain  at  the 
elastic  limit ;  thus  Weisbach  gives  EVJ  Ibs.  per  sq.  inch  for 
wrought  iron  ==  28,000,000=  double  the  E,.  for  cast  iron 
while  the  compressive  stresses  at  the  elastic  limit  are  the 
same  for  both  materials  (nearly). 

188.  General  Problem  of  Internal  Stress. — This,  as  treated 
in  the  mathematical  Theory  of  Elasticity,  developed  by 
Lame,  Clapeyron  and  Poisson,  may  be  stated  as  follows  : 

Given  the  original  form  of  a  body  when  free  from  stress, 
and  certain  co-efficients  depending  on  its  elastic  proper- 
ties ;  required  the  new  position,  the  altered  shape,  and  the  in- 
tensity of  the  stress  on  each  of  the  six  faces,  of  every  element 
of  the  body,  when  a  given  balanced  system  of  forces  is  applied 
to  the  body. 

Solutions,  by  this  theory,  of  certain  problems  of  the  na- 
ture just  given  involve  elaborate,  intricate,  and  bulky 
analysis ;  but  for  practical  purposes  Navier's  theories 
(1838)  and  others  of  more  recent  date,  are  sufficiently  exact, 
when  their  moduli  are  properly  determined  by  experiments 
covering  a  wide  range  of  cases  and  materials.  These  will 
be  given  in  the  present  work,  and  are  comparatively  sim- 
ple. In  some  cases  graphic  will  be  preferred  to  analytic 
methods  as  more  simple  and  direct,  and  indeed  for  some 
problems  they  are  the  only  methods  yet  discovered  for  ob- 
taining solutions.  Again,  experiment  is  relied  on  almost 
exclusively  in  dealing  with  bodies  of  certain  forms  under 
peculiar  systems  of  forces,  empirical  formulse  being  based 
on  the  experiments  made ;  e.g.,  the  collapsing  of  boilei 
flues,  and  in  some  degree  the  flexure  of  long  columns. 


206  MECHANICS   OF   ENGINEERING. 

189.  Classification  of  Cases. — Although,  in  almost  any  case 
whatever  of  the  deformation  of  a  solid  body  by  a  balanced 
system  of  forces  acting  on  it,  normal  and  shearing  stresses 
are  both  developed  in  every  element  which  is  affected  at 
all  (according  to  the  plane  section  considered,)  still,  cases 
where  the  body  is  prismatic,  and  the  external  system  con- 
sists of  two  equal  and  opposite  forces,  one  at  each  end  of 
the  piece  and  directed  away  from  each  other,  are  commonly 
called  cases  of  Tension;  (Fig.  192);  if  the  piece  is  a  short 
prism  with  the  same  two  terminal  forces  directed  toward 
each  other,  the  case  is  said  to  be  one  of  Compression ;  a  case 
similar  to  the  last,  but  where  the  prism  is  quite  long 
("  long  column  "),  is  a  case  of  Flexure  or  bending,  as  are  also 
most  cases  where  the  "applied  forces"  (i.e.,  the  external 
forces),  are  not  directed  along  the  axis  of  the  piece.   Rivet- 
ed joints  and  "  pin-connections  "  present  cases  of  Shearing; 
a  twisted  shaft  one  of  Torsion.     When  the  gravity  forces 
due  to  the  weights  of  the  elements  are  also  considered,  a 
combination  of  two  or  more  of  the  foregoing  general  cases 
may  occur. 

In  each  case,  as  treated,  the  principal  objects  aimed  at 
are,  so  to  design  the  piece  or  its  loading  that  the  greatest 
stress,  in  whatever  element  it  may  occur,  shall  not  exceed 
a  safe  value  ;  and  sometimes,  furthermore,  to  prevent  too 
great  deformation  on  the  part  of  the  piece.  The  first  ob- 
ject is  to  provide  sufficient  strength ;  the  second  sufficient 
stiffness.  / 

190.  Temperature  Stresses. — If  a  piece  is  under  such  con- 
straint that  it  is  not  free  to  change  its  form  with  changes 
of  temperature,  external  forces  are  induced,  the  stresses 
produced  by  which  are  called  temperature  stresses. 


TENSION. 


207 


TENSION. 

191.  Hooke's  Law  by  Experiment. — As  a  typical  experiment 
in  the  tension  of  a  long  rod  of  ductile  metal  such  as 
wrought  iron  and  the  mild  steels,  the  following  table  is  quot- 
ed from  Prof.  Cotterill's  "  Applied  Mechanics."  The  experi- 
ment is  old,  made  by  Hodgkinson  for  an  English  Eailway 
Commission,  but  well  adapted  to  the  purpose.  From  the 
great  length  of  the  rod,  which  was  of  wrought  iron  and 
0.517  in.  in  diameter,  the  portion  whose  elongation  was 
observed  being  49  ft.  2  in.  long,  the  small  increase  in  length 
below  the  elastic  limit  was  readily  measured.  The  succes- 
sive loads  were  of  such  a  value  that  the  tensile  stress 
p=P-:rF,  or  normal  stress  per  sq.  in.  in  the  transverse 
section,  was  made  to  increase  by  equal  increments  of  2657.5 
Ibs.  per  sq.  in.,  its  initial  value.  After  each  application  of 
load  the  elongation  was  measured,  and  after  the  removal 
of  the  load,  the  permanent  set,  if  any. 

TaDle  of  elongations  of  a  wrought  iron  rod,  of  a  leugth=49  ft.  2  in. 


p 

A 

Jl 

£=l+l 

X 

Load,  (Ibs.  per 
square  inch.) 

Elongation, 
(inches.) 

Increment 
of 
Elongation. 

e,  the  relative 
elongation,  (ab- 
stract number.) 

Permanent 
Set, 
(inchea) 

1X3667.5 

.0485 

.0485 

0.000082 

2X    " 

.1095 

.061 

.000186 

3X    " 

.1675 

.058 

.000283 

0.0019 

4X    " 

.224 

.0565 

.000379 

.002 

5X    " 

.2805 

.0565 

.000475 

.002? 

ex  " 

.337 

.0565 

.000570 

.003 

TX    " 

.393 

.056 

.004 

8X    " 

.452 

.059 

.000766 

.007S 

9X    " 

.5155 

.0635 

.0195 

10X    " 

.598 

.0825 

.049 

MX    " 

.760 

.162 

.1545 

12X    " 

1.810 

.550 

.667 

etc. 

208 


MECHANICS   OP   ENGINEERING. 


Eeferring  now  to  Fig.  198,  the  notation  is  evident.  P 
is  the  total  load  in  any  experiment,  F  the  cross  section  of 
the  rod  ;  hence  the  normal  stress  on  the  transverse  section 
is  p=P-r.F.  When  the  loads  are  increased  by  equal  in- 
crements, the  corresponding  increments  of  the  elongation 
A  should  also  be  equal  if  Hooke's  law  is  true.  It  will  be 
noticed  in  the  table  that  this  is  very  nearly  true  up  to  the 
8th  loading,  i.e.,  that  J^,  the  difference  between  two  con- 
secutive values  of  A,  is  nearly  constant.  In  other  words  the 
proposition  holds  good  : 


if  P  and  Pi  are  any  two  loads  below  the  8th,  and  /I  and  i, 
the  corresponding  elongations. 

The  permanent  set  is  just  perceptible  at  the  3d  load,  and 
increases  rapidly  after  the  8th,  as  also  the  increment  of 
elongation.  Hence  at  the  8th  load,  which  produces  a  ten- 
sile stress  on  the  cross  section  of  p=  8x2667.5=  21340.0 
Ibs.  per  sq.  inch,  the  elastic  limit  is  reached. 

As  to  the  state  of  stress  of  the  individual  elements,  if 
we  conceive  such  sub  -division 
of  the  rod  that  four  edges  of 
each  element  are  parallel  to  the 
axis  of  the  rod,  we  find  that  it 
is  in  equilibrium  between  two 
normal  stresses  on  its  end  faces 
(Fig.  199)  of  a  value  =pdF~ 
(P^F}dF  where  dF  is  the  hor- 
izontal section  of  the  element. 
If  dx  was  the  original  length, 
and  dh  the  elongation  produced  by  pdF,  we  shall  have, 
since  all  the  dx's  of  the  length  are  equally  elongated  at  the 
same  time, 

dl     X 


TENSION.  209 

where  1=  total  (original)  length.  But  dX-^dx  is  the  rela- 
tive elongation  e,  and  by  definition  (§  184)  the  Modulus  of 
Elasticity  for  Tension,  El}=p-r-e 


dx 

Eq.  (1)  enables  us  to  solve  problems  involving  the  elonga- 
tion of  a  prism  under  tension,  so  long  as  the  elastic  limit 
is  not  surpassed. 

The  values  of  E^  computed  from  experiments  like  those 
just  cited  should  be  the  same  for  any  load  under  the  elas- 
tic limit,  if  Hooke's  law  were  accurately  obeyed,  but  in 
reality  they  differ  somewhat,  especially  if  the  material 
lacks  homogeneity.  In  the  present  instance  (see  Table) 
we  have  from  the 

2d  Exper.  E=p+  6=28,680,000  Ibs.  per  sq.  in. 

5th     "        E,=  "    =28,009,000 

8th     "        Et=  "    =27,848,000        "  " 

If  similar  computations  were  made  beyond  the  elastic 
limit,  i.e.,  beyond  the  8th  Exper.,  the  result  would  be  much 
smaller,  showing  the  material  to  be  yielding  much  more 
readily. 

192.  Strain  Diagrams.  —  If  we  plot  the  stresses  per  sq.  inch 
(p)  as  ordinates  of  a  curve,  and  the  corresponding  relative 
elongations  (e)  as  abscissas,  we  obtain  a  useful  graphic  re- 
presentation of  the  results  of  experiment. 

Thus,  the  table  of  experiments  just  cited  being  utilized 
in  this  way,  we  obtain  on  paper  a  series  of  points  through 
which  a  smooth  curve  may  be  drawn,  viz.  :  OBC  Fig.  200, 
for  wrought  iron.  Any  convenient  scales  may  be  used  for 
p  and  e  ;  and  experiments  having  been  made  on  other 
metals  in  tension  and  the  results  plotted  to  the  same  scales 


210 


MECHANICS  OF   ENGINEERING. 


as  before  for  p  and  e,  we  have  the  means  of  comparing  their 
tensile  properties.  Fig.  200  shows  two  other  curves,  rep- 
resenting (roughly)  the  average  behavior  of  steel  and  cast 
iron.  At  the  respective  elastic  limits  J?,  B  ,  and  B ',  it  will 
be  noticed  that  the  curve  for  wrought  iron  makes  a  sudden 
turn  from  the  vertical,  while  those  of  the  others  curve  away 
more  gradually ;  that  the  curve  for  steel  lies  nearer  the 
vertical  axis  than  the  others,  which  indicates  a  higher 
value  for  Et ;  and  that  the  ordinates  BA',  B'A',  and  B  'A  " 
(respectively  21,000,  9,000,  and  30,000  Ibs.  per  sq.  inch)  in- 


dicate  the  tensile  stress  at  the  elastic  limit.  These  latter 
quantities  will  be  called  the  moduli  of  tenacity  at  elastic 
limit  for  the  respective  materials.  [On  a  true  scale  the 
point  C  would  be  much  further  to  the  right  than  here 
shown.  Only  one  half  of  the  curve  for  steel  is  given,  for 
want  of  space.] 

Within  the  elastic  limit  the  curves  are  nearly  straight 
(proving  Hooke's  law)  and  if  a,  a,  and  a"  are  the  angles 
made  by  these  straight  portions  with  the  axis  of  X  (i.e., 
of  e),  we  shall  have 


(Et  for  w.  iron) :  (Et  c.  iron) :  (Et  steel) : :  tan  a :  tan  a' :  tan  a" 


TENSION.  211 

as  a  graphic  relation  between  their  moduli  of  elasticity 
(since  Eu=^-). 

Beyond  the  elastic  limit  tins  wrought  iron  rod  shows  large 
increments  of  elongation  for  small  increments  of  stress, 
i.e.,  the  curve  becomes  nearly  parallel  to  the  horizontal 
axis,  until  rupture  occurs  at  a  stress  of  53,000  Ibs.  per  sq. 
inch  of  original  sectional  area  (at  rupture  this  area  is  some- 
what reduced,  especially  in  the  immediate  neighborhood 
of  the  section  of  rupture ;  see  next  article)  and  after  a  rel- 
ative elongation  e=  about  0.30,  or  30%.  (The  preceding 
table  shows  only  a  portion  of  the  results.)  The  curve 
for  steel  shows  a  much  higher  breaking  stress  (100,000 
Ibs.  per  sq.  in.)  than  the  wrought  iron,  but  the  total 
elongation  is  smaller,  e=  about  10%.  This  is  an  average 
curve ;  tool  steels  give  an  elongation  at  rupture  of  about 
4  to  5%,  while  soft  steels  resemble  wrought  iron  in  their 
ductility,  giving  an  extreme  elongation  of  from  10  to  20%. 
Their  breaking  stresses  range  from  70,000  to  150,000  Ibs. 
or  more  per  sq.  inch.  Oast  iron,  being  comparatively  brit- 
tle, reaches  at  rupture  an  elongation  of  only  3  or  4  tenths 
of  one  per  cent.,  the  rupturing  stress  being  about  18,000 
Ibs.  per  sq.  inch.  The  elastic  limit  is  rather  ill  denned  in 
the  case  of  this  metal ;  and  the  proportion  of  carbon  and 
the  mode  of  manufacture  have  much  influence  on  its  be- 
havior under  test. 

193.  Lateral  Contraction. — In  the  stretching  of  prisms  of 
nearly  all  kinds  of  material,  accompanying  the  elongation 
of  length  is  found  also  a  diminution  of  width  whose  rela- 
tive amount  in  the  case  of  the  three  metals  just  treated  ia 
about  y^  or  ^  of  the  relative  elongation  (within  elastic 
limit).  Thus,  in  the  third  experiment  in  the  table  of  §  191, 
this  relative  lateral  contraction  or  decrease  of  diameter 
=  yz  to  y±  of  s,  i.e.,  about  0.00008.  In  the  case  of  cast 
iron  and  hard  steels  contraction  is  not  noticeable  ex- 


212  MECHANICS   OF   ENGINEERING 

cept  by  very  delicate  measurements,  both  within  and  with- 
out the  elastic  limit ;  but  the  more  ductile  metals,  as 
wrought  iron  and  the  soft  steels,  when  stretched  beyond 
the  elastic  limit  show  this  feature  of  their  deformation 
in  a  very  marked  degree.  Fig.  201  shows  by  dotted  lines 
the  original  contour  of  a  wrought  iron  rod,  while  the  con- 
tinuous lines  indicate  that  at  rupture.  At  the  cross  section 
of  rupture,  whose  position  is  determined  by  some 
local  weakness,  the  drawing  out  is  peculiarly 
pronounced. 

The  contraction  of  area  thus  produced  is  some- 
times as  great  as  50  or  60%  at  the  fracture. 

194.  "Flow  of  Solids." — When  the  change  in  re- 
lative position  of  the  elements  of  a  solid  is  ex- 
treme, as  occurs  in  the  making  of  lead  pipe, 
drawing  of  wire,  the  stretching  of  a  rod  of  duc- 
tile metal  as  in  the  preceding  article,  we  have 
FIG.  201.     instances  of  what  is  called  the  Flow  of  Solids,  in- 
teresting  experiments    on    which    have    been    made   by 
Tresca. 

195.  Moduli  of  Tenacity. — The  tensile  stress  per  square 
inch  (of  original  sectional  area)  required  to  rupture  a 
prism  of  a  given  material  will  be  denoted  by  T  and  called 
the  modulus  of  ultimate  tenacity  ;  similarly,  the  modulus  of 
safe  tenacity,  or  greatest  safe  tensile  stress  on  an  element, 
by  T';  while  the  tensile  stress  at  elastic  limit  may  be 
called  T".  The  ratio  of  T  to  T"  is  not  fixed  in  practice 
but  depends  upon  circumstances  (from  %  to  ^). 

Hence,  if  a  prism  of  any  material  sustains  a  total  pull 
or  load  P,  and  has  a  sectional  area  =F,  we  have 

P=  FT  for  the  ultimate  or  breaking  load.  \ 

p=FT'   "      "    safe  load.  V     .    .     (2) 

p»=FT""      "     load  at  elastic  limit.  ) 

Of  course  T'  should  always  be  less  than  T". 


TENSION.  213 

196.  Resilience  of  a  Stretched  Prism.  —  Fig.  202.  In  the 
gradual  stretching  of  a  prism,  fixed  at  one  extremity,  the 
value  of  the  tensile  force  P  at  the  other  necessarily  de- 
pends on  the  elongation  A  at  each  stage  of  the  lengthening, 
according  to  the  relation  [eq.  (1)  of  §  191.] 


within  the  elastic  limit.  (If  we  place  a  weight  G  on  the 
flanges  of  the  unstretched  prism  and  then  leave 
it  to  the  action  of  gravity  and  the  elastic  action 
of  the  prism,  the  weight  begins  to  sink,  meeting 
an  increasing  pressure  P,  proportional  to  X,  from 
the  flanges).  Suppose  the  stretching  to  continue 
until  P  reaches  some  value  P"  (at  elastic  limit 
say),  and  X  a  value  A".  Then  the  work  done  so 
<J>+  far  is 

FII  202.         U=  mean  force  x  8Pace  -  #  P   r     '     '     (4> 
But  from  (2)  F  =  FT',  and  (see  §§  184  and  191) 


.-.  (4)  becomes  U=y2  T  e".  Fl=%  T"  e"  V     .    .     (5) 

where  V  is  the  volume  of  the  prism.  The  quantity  ^  T"s", 
or  work  done  in  stretching  to  the  elastic  limit  a  cubic 
inch  (or  other  unit  of  volume)  of  the  given  material,  Weis- 
bach  calls  the  Modulus  of  Resilience  for  tension.  From  (5) 
it  appears  that  the  amounts  of  work  done  in  stretching  to 
the  elastic  limit  prisms  of  the  same  material  but  of  difl'er- 
ent  dimensions  are  proportional  to  their  volumes  simply. 
The  quantity  %T"t"  is  graphically  represented  by  the 
area  of  one  of  the  triangles  such  as  OA'B,  OA'B"  in  Fig. 
200  ;  for  (in  the  curve  for  wrought  iron  for  instance)  the 
modulus  of  tenacity  at  elastic  limit  is  represented  by  A'  B, 
and  e"  (i.e.,  e  for  elastic  limit)  by  OA'.  The  remainder  of 


214  MECHANICS   OF    ENGINEERING. 

the  area  OBC  included  between  the  curve  and  the  hori- 
zontal axis,  i.e.,  from  B  to  C,  represents  the  work  done  in 
stretching  a  cubic  unit  from  the  elastic  limit  to  the  point 
of  rupture,  for  each  vertical  strip  having  an  altitude  =p 
and  a  width  =ds,  has  an  area  =pde,  i.e.,  the  work  done  by 
the  stress  p  on  one  face  of  a  cubic  unit  through  the  dis- 
tance ds,  or  increment  of  elongation. 

If  a  weight  or  load  =  O  be  "  suddenly  "applied  to  stretch 
the  prism,  i.e.,  placed  on  the  flanges,  barely  touching 
them,  and  then  allowed  to  fall^  when  it  comes  to  rest  again 
it  has  fallen  through  a  height  ^,  and  experiences  at  this 
instant  some  pressure  Pi  from  the  flanges;  PI=?  The 
work  Gh  has  been  entirely  expended  in  stretching  the 
prism,  none  in  changing  the  kinetic  energy  of  G,  which 
=0  at  both  beginning  and  end  of  the  distance  ^. 


Since  Pi=%Gr,  i.e.,  is  >  6?,  the  weight  does  not  remain  in 
this  position  but  is  pulled  upward  by  the  elasticity  of  the 
prism.  In  fact,  the  motion  is  harmonic  (see  §§  59  and 
138).  Theoretically,  the  elastic  limit  not  being  passed,  the 
oscillations  should  continue  indefinitely. 

Hence  a  load  G  "  suddenly  applied  "  occasions  double  the 
tension  it  would  if  compelled  to  sink  gradually  by  a  sup- 
port underneath,  which  is  not  removed  until  the  tension  is 
just  =  G,  oscillation  being  thus  prevented. 

If  the  weight  G  sinks  through  a  height  =h  before  strik- 
ing the  flanges,  Fig.  202,  we  shall  have  similarly,  within 
elastic  limit,  if  ^=  greatest  elongation,  (the  mass  of  rod 
being  small  compared  with  that  of  G). 

G(h+^^y2P^       '.        .        .        .    (6) 

If  the  elastic  limit  is  to  be  just  reached  we  have  from  eqs. 
(5)  and  (6),  neglecting  ^  compared  with  h, 

F         .        ',  .    ,  .        (7) 


TENSION. 


215 


an  equation  of  condition  that  the  prism  shall  not  be  in- 
jured. 

Example. — If  a  steel  prism  have  a  sectional  area  of  ^ 
6q.  inch  and  a  length  1=10  ft.  =120  inches,  what  is  the 
greatest  allowable  height  of  fall  of  a  weight  of  200  Ibs., 
that  the  final  tensile  stress  induced  may  not  exceed  T"= 
30,000  Ibs.  per  sq.  inch,  if  e"  =.002  ?  From  (7),  using  the 
inch  and  pound,  we  have 


197.  Stretching  of  a  Prism  by  Its  Own  Weight. — In  the  case 
of  a  very  long  prism  such  as  a  mining- 
pump  rod,  its  weight  must  be  taken  into 
account  as  well  as  that  of  the  terminal 
load  P ,  see  Fig.  203.  At  (a.)  the  prism 
is  shown  in  its  unstrained  condition ;  at 
(6)  strained  by  the  load  Pl  and  its  own 
weight.  Let  the  cross  section  be  =F,  the 
heaviness  of  the  prism  =;-.  Then  the  rela- 
tive extension  of  any  element  at  a  distance 


0    A 

(«•)  P 

FIG.  208. 

j  from  o  is 


(1) 


(See  eq.  (1)  §  191) ;  since  P^-\-F?x  is  the  load  hanging  upon 
the  cross  section  at  that  locality.  Equal  dx's,  therefore, 
are  unequally  elongated,  x  varying  from  0  to  I.  The  total 
elongation  is 


_      P.*     I    72  ur» 

~  FEt    1FE, 


Le.,  A=  the  amount  due  to  P,,  plus  an  extension  which 
half  the  weight  of  the  prism  would  produce,  hung  at  the 
lower  extremity. 


216  MECHANICS   OF   ENGINEERING. 

The  foregoing  relates  to  the  deformation  of  the  piece, 
and  is  therefore  a  problem  of  stiffness.  As  to  the  strength 
of  the  prism,  the  relative  elongation  £=ctt.-±-dx  [see  eq.  (1)], 
which  is  variable,  must  nowhere  exceed  a  safe  value  e'= 
I  '  +EL  (from  eq.  (1)  §  191,  putting  P=FT  ,  and  /=/). 
Now  the  greatest  value  of  the  ratio  dX  :  dx,  by  inspecting 
eq.  (1),  is  seen  to  be  at  the  upper  end  where  x=l.  The 
proper  cross  section  F,  for  a  given  load  Plt  is  thus  found. 


Putting         j  we  have  p  =L_          m        (2) 


198.  Solid  of  Uniform  Strength  in  Tension,  or  Twinging  body 
of  minimum  material  supporting  its  own 
weight  and  a  terminal  load  P,.  Let  it  be  a 
solid  of  revolution.  If  every  cross-section 
F  at  a  distance  —x  from  the  lower  extrem- 
ity, bears  its  safe  load  FT',  every  element 
of  the  body  is  doing  full  duty,  and  its  form 
is  the  most  economical  of  material. 

The   lowest  section  must  have  an  area 
FIG.  204.         F0=P^  T',  since  P,  is  its  safe  load.     Fig. 
204.     Consider  any  horizontal  lamina  ;  its  weight  is  fFdx, 
(f=  heaviness  of  the  material,  supposed  homogenous),  and 
its  lower  base  F  must  have  Pj+  G  for  its  safe  load,  i.e. 

G+P^FT'        .        .      :,          (1) 

in  which  G  denotes  the  weight  of  the  portion  of  the  solid 
below  F.     Similarly  for  the  upper  base  F+dF,  we  have 


.        .        (2) 
By  subtraction  we  obtain 

rFdx=T'dF',    i.e.  lcte=    ** 
1  F 


TENSION.  217 

in  which  the  two  variables  x  and  F  are  separated.     By  in- 
tegration we  now  have 


(3) 


i.e.,  F=FQer  =j±  er'       (4) 

from  which  J^may  be  computed  for  any  value  of  x. 

The  weight  of  the  portion  below  any  F  is  found  from  (1) 
and  (4) ;  i.e. 


while  the  total  extension  ^  will  be 

rrri 

l=*"~l (6) 

the  relative  elongation  dX-^-dx  being  the  same  for  every  dx 
and  bearing  the  same  ratio  to  e"  (at  elastic  limit),  as  T' 
does  to  T". 

199.  Tensile  Stresses  Induced  by  Temperature.— If  the  two 
ends  of  a  prism  are  immovably  fixed,  when  under  no  strain 
and  at  a  temperature  t,  and  the  temperature  is  then  low- 
ered to  a  value  t't  the  body  suffers  a  tension  proportional 
to  the  fall  in  temperature  (within  elastic  limit).  If  for  a 
rise  or  fall  of  1°  Fahr.  (or  Cent.)  a  unit  of  length  of  the 
material  would  change  in  length  by  an  amount  7  (called 
the  co-efficient  of  expansion)  a  length  =1  would  be  con- 
tracted an  amount  ),=ftl(t-t'}  during  the  given  fall  of  tem- 
perature if  one  end  were  free.  Hence,  if  this  contraction 
is  prevented  by  fixing  both  ends,  the  rod  must  be  under  a 
tension  P,  equal  in  value  to  the  force  which  would  be 


218  MECHANICS   OF  ENGINEERING. 

necessary  to  produce  the  elongation  /,  just  stated,  under 
ordinary  circumstances  at  the  lower  temperature. 

From  eq.  (1)  §191,  therefore,  we  have  for  this  tension 
due  to  fall  of  temperature 


P= 


For  1°  Cent,  we  may  write 

For  Cast  iron  y  =  .0000111  ; 
"  Wrought  iron  —  .0000120  ; 
"  Steel  =  .0000108  to  .0000114  ; 

"     Copper          rj  =  .0000172  ; 
"     Zinc  =  .0000300. 


COMPRESSION    OF    SHORT    BLOCKS. 

200.  Short  and  Long  Columns. — In  a  prism  in  tension,  its 
own  weight  being  neglected,  all  the  elements  between  the 
localities  of  application  of  the  pair  of  external  forces  pro- 
ducing the  stretching  are  in  the  same  state  of  stress,  if  the 
external  forces  act  axially  (excepting  the  few  elements  in  the 
immediate  neighborhood  of  the  forces ;  these  suffering 
local  stresses  dependent  on  the  manner  of  application  of 
the  external  forces),  and  the  prism  may  be  of  any  length 
without  vitiating  this  statement.  But  if  the  two  external 
forces  are  directed  toward  each  other  the  intervening  ele- 
ments will  not  all  be  in  the  same  state  of  compressive 
stress  unless  the  prism  is  comparatively  short  (or  unless 
numerous  points  of  lateral  support  are  provided).  A  long 
prism  will  buckle  out  sideways,  thus  even  inducing  tensile 
stress,  in  some  cases,  in  the  elements  on  the  convex  side. 

Hence  the  distinction  between  short  blocks  and  long 
columns.  Under  compression  the  former  yield  by  crush- 
ing or  splitting,  while  the  latter  give  way  by  flexurp  fi-e' 
bending).  Long  columns,  then  will  be  treated  separately 


COMPRESSION   OF    SHORT   BLOCKS. 

in  a  subsequent  chapter.  In  the  present  section  the  blocks 
treated  being  about  three  or  four  times  as  long  as  wide, 
all  the  elements  will  be  considered  as  being  under  equal 
compressive  stresses  at  the  same  time. 

201.  Notation  for  Compression. — By  using  a  subscript  c, 
we  may  write 

JEC=  Modulus   of  Elasticity;*    i.e.    the    quotient  of   the 
compressive  stress  per  unit  of  area  divided  by  the  relative " 
shortening ;  also 

C=  Modulus  of  crushing ;  i.e.  the  force  per  unit  of  sec- 
tional area  necessary  to  rupture  the  block  by  crushing ; 

C'=  Modulus  of  safe  compression,  a  safe  compressive 
stress  per  unit  of  area ;  and 

C"=  Modulus  of  compression  at  elastic  limit. 

For  the  absolute  and  relative  shortening  in  length  we 
may  still  use  A  and  e,  respectively,  and  within  the  elastic 
limit  may  write  equations  similar  to  those  for  tension,  F 
being  the  sectional  area  of  the  block  and  P  one  of  the  ter- 
minal forces,  while  p  =  compressive  stress  per  unit  of  area 
of  F,  viz.: 


within  the  elastic  limit. 
Also  for  a  short  block 

Crushing  force  =FC 

Compressive  force  at  elastic  limit  —FG"     \   .     (2) 

Safe  compressive  force  =FC' 

202.  Remarks  on  Crushing. — As  in  §  182  for  a  tensile 
stress,  so  for  a  compressive  stress  we  may  prove  th-at  a 

*  [NOTE.— It  must  be  remembered  that  the  modulus  of  elasticity, 
whether  for  normal  or  shearing  stresses,  is  a  number  indicative  of  stiff- 
ness, not  of  strength,  and  has  no  relation  to  the  elastic  limit  (except 
that  experiments  to  determine  it  must  not  pass  that  limit).] 


220  MECHANICS   OF  ENGINEERING. 

shearing  stress  =p  sin  a  cos  a  is  produced  on  planes  at  an 
angle  a  with  the  axis  of  the  short  block,  p  being  the  com- 
pression per  unit  of  area  of  transverse  section.  Accord- 
ingly it  is  found  that  short  blocks  of  many  comparatively 
brittle  materials  yield  by  shearing  on  planes  making  an 
angle  of  about  45°  with  the  axis,  the  expression  p  sin  a 
cos  a  reaching  a  maximum,  for  a=45° ;  that  is,  wedge- 
shaped  pieces  are  forced  out  from  the  sides.  Hence  the 
necessity  of  making  the  block  three  or  four  times  as  long 
as  wide,  since  otherwise  the  friction  on  the  ends  would 
cause  the  piece  to  show  a  greater  resistance  by  hindering 
this  lateral  motion.  Crushing  by  splitting  into  pieces 
parallel  to  the  axis  sometimes  occurs. 

Blocks  of  ductile  material,  however,  yield  by  swelling 
out,  or  bulging,  laterally,  resembling  plastic  bodies  some- 
what in  this  respect. 

The  elastic  limit  is  more  difficult  to  locate  than  in  ten- 
sion, but  seems  to  have  a  position  corresponding  to  that 
in  tension,  in  the  case  of  wrought  iron  and  steel.  With 
cast  iron,  however,  the  relative  compression  at  elastic 
limit  is  about  double  the  relative  extension  (at  elastic 
limit  in  tension),  but  the  force  producing  it  is  also  double. 
For  all  three  metals  it  is  found  that  Ec=Et  quite  nearly, 
so  that  the  single  symbol  E  may  be  used  for  both. 


EXAMPLES  IN  TENSION  AND  COMPRESSION. 

203.  Tables  for  Tension  and  Compression. — The  round  num- 
bers of  the  following  tables  are  to  be  taken  as  rude  averages 
only,  for  use  in  the  numerical  examples  following.  (The 
scope  and  design  of  the  present  work  admit  of  nothing 
more.  For  abundant  detail  of  the  results  of  the  more  im- 
portant experiments  of  late  years,  the  student  is  referred 
to  the  recent  works 'of  Profs.  Thurston,  Burr,  Lanza,  and 
Wood).  Another  column  might  have  been  added  giving 
the  Modulus  of  ^Resilience  in  each  case,  viz.:  y2  e"T" 

/77"2\ 

(which  also   =—  ^  )  ;     see  §  196.     e  is  an  abstract  num- 

2  At  / 


EXAMPLES  IN  TENSION  AND   COMPRESSION.         221 


her,  and  =X-±l,  while  Ett  T",  and  T  are  given  in  pounds 
per  square  inch: 

TABLE   OF  THE   MODULI,    ETC.,    OF   MATERIALS    IN   TENSION. 


e" 

e 

% 

T" 

T 

Material. 

(Elastic  limit.) 

At  Rupture. 

Mod.  of  Blast. 

Elastic  limit. 

Rupture. 

abst.  number. 

abst.  number. 

bs.  per  sq.  in. 

Ibs.  per  sq.  in. 

bs.  per  sq.  in. 

Soft  Steel, 

.00200 

.2500 

26,000,000 

50,000 

80,000 

Hard  Steel, 

.00200 

.0500 

40,000,000 

99,000 

130,000 

Cast  Iron, 
Wro't  Iron, 

Brass, 

.00066 
.00080 

.00100 

.0020 
.2500 

14,000,000 
28,000,000 

10,000,000 

9,000 
22,000 

(   7,000 
\     to 
j  19,000 

18,000 
(45,000 
\    to 
(60,000 
16.000 
to 
50,000 

Glass, 

9,000,000 

3,500 

Wood,    with 
the  fibres, 

1        .00200 
<          to 
.01100 

.0070 
to 
.0150 

200,000 
to 
2,000,000 

3,000 
to 
19,000 

6,000 
to 

28.000 

Hemp  rope, 

7,000 

[N.B.— Expressed  in  kilograms  per  square  centim.,  Et,  T  and  T"    would  be  nu 
merically  about  V]4  as  large  as  above,  while  e  and  e"  would  be  unchanged.] 

TABLE   OF   MODULI,  ETC.;   COMPRESSION   OF   SHORT   BLOCKS. 


e" 

e 

Ec 

C" 

C 

Material. 

Elastic  limit. 

At  lupture. 

Mod.  of  Blast. 

Elastic  limit. 

Rupture. 

abst.  number 

abst.  number. 

Ibs.  per  sq.  in. 

Ibs.  per  sq.  in. 

Ibs.  per  sq.  in. 

Soft  Steel, 

0.00100 

30,000,000 

30,000 

Hard  Steel. 

0.00120 

0.3000 

40,000,000 

50,000 

200,000 

Cast  Iron, 

0.00150 

14,000.000 

20,000 

90,000 

Wro't  Iron, 

0.00080 

0.3000 

28,000,000 

24,000 

40,000 

Glass, 

20,000 

Granite, 
Sandstone, 

|    See 

10,000 
5,000 

Brick, 

J 

3,000 

Wood,  with 
the  fibres, 

(   0.0100 
•<      to 
(   0.0400 

350,000 
2,000,000 

2,000 
to 
10,000 

Portland     / 
Cement,  f 

(I  813a) 

4,000 

222  MECHANICS   OF   ENGINEERING. 

204.  Examples.  No.  1.—  A  bar  of  tool  steel,  of  sectional 
area  =0.097  sq.  inches,  is  ruptured  by  a  tensile  force  of 
14,000  Ibs.  A  portion  of  its  length,  originally  ^  a  foot, 
is  now  found  to  have  a  length  of  0.532  ft.  Required  T, 
and  e  at  rupture.  Using  the  inch  and  pound  as  units  (as 
in  the  foregoing  tables)  we  have  jT=y^?=144326  Ibs.  per 
sq.  in.;  (eq.  (2)  §  195)  ;  while 

e=(0.532—  0.5)xl2-=-(0.50xl2)=0.064. 

EXAMPLE  2.—  Tensile  test  of  a  bar  of  "  Hay  Steel  "  for 
the  Glasgow  Bridge,  Missouri.  The  portion  measured  was 
originally  3.21  ft.  long  and  2.09  in.  X  1.10  in.  in  section. 
At  the  elastic  limit  P  was  124,200  Ibs.,  and  the  elongation 
was  0.064  ins.  Required  Elt  T",  and  e"  (for  elastic  limit). 

"  =.00165  at  elastic  limit. 


_ 
2"'=124,200H-(2.09x  1.10)=  54,000  Ibs.  per  sq.  in. 

lbs-  er  ••  in 


Nearly  the  same  result  for  Et  would  probably  have  been 
obtained  for  values  of  p  and  e  below  the  elastic  limit. 

The  Modulus  of  Resilience  of  the  above  steel  (see  §  196) 
would  be  %  s"  T  "=44.82  inch-pounds  of  work  per  cubic 
inch  of  metal,  so  that  the  whole  work  expended  in  stretch- 
ing to  the  elastic  limit  the  portion  above  cited  is 

#=  ^  e"  T"  F=3968.  inch  -lbs. 

An  equal  amount  of  work  will  be  done  by  the  rod  in  re- 
covering its  original  length. 

EXAMPLE  3.  —  A  hard  steel  rod  of  ^  sq.  in.  section  and 
20  ft.  long  is  under  no  stress  at  a  temperature  of  130° 


EXAMPLES  IN  TENSION   AND   COMPRESSION.        223 

Cent.,  and  is  provided  with  flanges  so  that  the  slightest 
contraction  of  length  will  tend  to  bring  two  walls  nearer 
together.  If  the  resistance  to  this  motion  is  10  tons  how 
low  must  the  temperature  fall  to  cause  any  motion  ?  i)  be- 
ing =  .0000120  (Cent,  scale).  From  §  199  we  have,  ex- 
pressing P  in  Ibs.  and  F  in  sq.  inches,  since  Et=  40,000,000 
(6s.  per  sq.  inch, 

10x2,000=40,000,000  x  #  X  (130-*')  X  0.000012;  whence 
r=46.6°  Centigrade. 

EXAMPLE  4. — If  the  ends  of  an  iron  beam  bearing  5  tons 
at  its  middle  rest  upon  stone  piers,  required  the  necessary 
bearing  surface  at  each  pier,  putting  C'  for  stone  =200 
Ibs.  per  sq.  inch.  25  sq.  in.,  Ans. 

EXAMPLE  5. — How  long  must  a  wrought  iron  rod  be, 
supported  vertically  at  its  upper  end,  to  break  with  its 
own  weight  ?  216,000  inches,  Ans. 

EXAMPLE  6. — One  voussoir  (or  block)  of  an  arch -ring 
presses  its  neighbor  with  a  force  of  50  tons,  the  joint  hav- 
ing a  surface  of  5  sq.  feet ;  required  the  compression  per 
sq.  inch.  138.8  Ibs.  per  sq.  in.,  Ans. 

205.  Factor  of  Safety. — When,  as  in  the  case  of  stone,  the 
value  of  the  stress  at  the  elastic  limit  is  of  very  uncertain 
determination  by  experiment,  it  is  customary  to  refer  the 
value  of  the  safe  stress  to  that  of  the  ultimate  by  making 
it  the  w'th  portion  of  the  latter,  n  is  called  a  factor  of 
safety,  and  should  be  taken  large  enough  to  make  the  safe 
stress  come  within  the  elastic  limit.  For  stone,  n  should 
not  be  less  than  10,  i.e.  (7/=(7-i-w;  (see  Ex.  6,  just  given). 


206.  Practical  Notes.— It  was  discovered  independently  by 
Commander  Beardslee  and  Prof.  Thurston,  in  1873,  that 
if  wrought  iron  rods  were  strained  considerably  beyond 
the  elastic  limit  and  allowed  to  remain  free  from  stress 


224  MECHANICS   OF    ENGINEERING. 

for  at  least  one  day  thereafter,  a  second  test  would  show 
higher  limits  both  elastic  and  ultimate. 

When  articles  of  cast  iron  are  imbedded  in  oxide  of  iron 
and  subjected  to  a  red  heat  for  some  days,  the  metal  loses 
most  of  its  carbon,  and  is  thus  nearly  converted  into 
wrought  iron,  lacking,  however,  the  property  of  welding. 
JBeing  malleable,  it  is  called  malleable  cast  iron. 

Chrome  steel  (iron  and  chromium)  and  tungsten  steel  pos- 
sess peculiar  hardness,  fitting  them  for  cutting  tools,  rock 
drills,  picks,  etc. 

By  fatigue  of  metals  we  understand  the  fact,  recently  dis- 
covered by  Wohler  in  experiments  made  for  the  Prussian 
Government,  that  rupture  may  be  produced  by  causing  the 
stress  on  the  elements  to  vary  repeatedly  between  two 
limiting  values,  the  highest  of  which  may  be  considerably 
below  T  (or  (7),  the  number  of  repetitions  necessary  to 
produce  rupture  being  dependent  both  on  the  range  of 
variation  and  the  higher  value. 

For  example,  in  the  case  of  Phoenix  iron  in  tension, 
rupture  was  produced  by  causing  the  stress  to  vary  from 
)  to  52,800  Ibs.  per  sq.  inch,  800  times ;  also,  from  0  to 
<4,000  Ibs.  per  sq.  inch  240,853  times  ;  while  4,000,000  va- 
riations between  26,400  and  48,400  per  sq.  inch  did  not 
cause  rupture.  Many  other  experiments  were  made  and 
the  following  conclusions  drawn  (among  others): 

Unlimited  repetitions  of  variations  of  stress  (Ibs.  per 
yjq.  in.)  between  the  limits  given  below  will  not  injure  the 
metal  (Prof.  Burr's  Materials  of  Engineering). 

Wrought  iron    I  From  17'600  ComP-  to  17,600%Tension. 
'    (      "  0  to  33,000 

(  From  30,800  Comp.  to  30,800  Tension. 
Axle  Cast  Steel.-]      "  0  to  52,800 

(     "       38500  Tens,    to  88,000 
(See  p.  232  for  an  addendum  to  this  paragraph.) 


SHEARING. 

SHEARING. 


225 


207.  Rivets. — The  angular  distortion  called  shearing 
strain  in  the  elements  of  a  body,  is  specially  to  be  provided 
for  in  the  case  of  rivets  joining  two  or  more  plates.  This 
distortion  is  shown,  in  Figs.  205  and  206,  in  the  elements 
near  ';he  plane  of  contact  of  the  plates,  much  exaggerated. 


FIG.  206. 


In  Fig.  205  (a  lap-joint)  the  rivet  is  said  to  be  in  single 
shear ;  in  Fig.  206  in  double  shear.  If  P  is  just  great 
enough  to  shear  off  the  rivet,  the  modulus  of  ultimate  shear- 
ing, which  may  be  called  S,  (being  the  shearing  force  per 
unit  of  section  when  rupture  occurs)  is 


(1) 


in  which  F=  the  cross  section  of  the  rivet,  its  diameter 
being  =d.  For  safety  a  value  >S"=  ]/^  to  y£>  of  S  should 
be  taken  for  metal,  in  order  to  be  within  the  elastic  limit. 

As  the  width  of  the  plate  is  diminished  by  the  rivet 
hole  the  remaining  sectional  area  of  the  plate  should  be 
ample  to  sustain  the  tension  P,  or  2P,  (according  to  the 
plate  considered,  see  Fig.  206),  P  being  the  safe  shearing 
force  for  the  rivet.  Also  the  thickness  t  of  the  plate 
should  be  such  that  the  side  of  the  hole  shall  be  secure 
against  crushing  ;  P  must  not  be  >  C'td,  Fig.  205. 

Again,  the  distance  a,  Fig.  205,  should  be  such  as  to 
prevent  the  tearing  or  shearing  out  of  the  part  of  the 
plate  between  the  rivet  and  edge  of  the  plate. 


226  MECHANICS  OF   ENGINEERING. 

For  economy  of  material  the  seam  or  joint  should  be 
no  more  liable  to  rupture  by  one  than  by  another,  of  the 


four  modes  just  mentioned.  The  relations  which  must 
then  subsist  will  be  illustrated  in  the  case  of  the  "  butt- 
joint  "  with  two  cover-plates,  Fig.  207.  Let  the  dimen- 
sions be  denoted  as  in  the  figure  and  the  total  tensile  force 
on  the  joint  be  =  Q.  Each  rivet  (see  also  Fig.  206)  is  ex- 
posed in  each  of  two  of  its  sections  to  a  shear  of  ^  Q, 
hence  for  safety  against  shearing  of  rivets  we  put 


(1) 


Along  one  row  of  rivets  in  the  main  plate  the  sectional 
area  for  resisting  tension  is  reduced  to  (b  —  3d)tlt  hence  for 
safety  against  rupture  of  that  plate  by  the  tension  Q,  we 
put 

'      ......     (2) 


Equations  (1)  and  (2)  suffice  to  determine  d  for  the  rivets 
and  t{  for  the  main  plates,  Q  and  b  being  given;  but  the 
values  thus  obtained  should  also  be  examined  with  refer- 
ence to  the  compression  in  the  side  of  the  rivet  hole,  i.e., 
yh  Q  must  not  be  >  C'ttd.  [The  distance  a,  Fig.  205,  to  the 
edge  of  the  plate  is  recommended  by  different  authorities 
to  be  from  d  to  3d] 

Similarly,  for  the  cover-plate  we  must  have 


.        .        .     (3) 

< 
and  /2Qnot  >  C'td~ 


SHEARING.  227 

If  the  rivets  do  not  fit  their  holes  closely,  a  large  margin 
should  be  allowed  in  practice.  Again,  in  boiler  work,  the 
pitch,  or  distance  between  centers  of  two  consecutive  rivets 
may  need  to  be  smaller,  to  make  the  joint  steam-tight,  than 
would  be  required  for  strength  alone. 

208.  Shearing  Distortion. — The  change  of  form  in  an  ele- 
ment due  to  shearing  is  an  angular  deformation  and  will 
be  measured  in  /r-measure.  This  angular  change  or  dif- 
ference between  the  value  of  the  corner  angle  during  strain 
and  *-£TT,  its  value  before  strain,  will  be  called  d,  and  is 
proportional  (within  elastic  limit)  to  the  shearing  stress 
per  unit  of  area,  ps,  existing  on  all  the  four  faces  whose 
angles  with  each  other  have  been  changed. 

Fig.  208.  (See  §  181).  By  §  184  the  Modulus  of  Shearing 
Elasticity  is  the  quotient  obtained  by  dividing  pK  by  d  ;  i.e. 
(elastic  limit  not  passed), 


.....    (1) 

or  inversely,  S=ps+Es (1)' 

The  value  of  E&  for  different  substances  is  most  easily 
determined  by  experiments  on  torsion 
in  which  shearing  is  the  most  promi- 
nent stress.  (This  prominence  depends 
on  the  position  of  the  bounding  planes 
of  the  element  considered ;  e.g.,  in  Fig. 

]/_ i_  208,  if  another  element  were  considered 

within  the  one  there  shown  and  with 
FIG.  908.  its  planes  at  45°  with  those  of  the  first, 

we  should  find  tension  alone  on  one  pair  of  opposite  faces, 
compression  alone  on  the  other  pair.)  It  will  be  noticed 
that  shearing  stress  cannot  be  present  on  two  opposite 
faces  only,  but  exists  also  on  another  pair  of  faces  (those 
perpendicular  to  the  stress  on  the  first),  forming  a  couple 
of  equal  and  opposite  moment  to  the  first,  this  being 
necessary  for  the  equilibrium  of  the  element,  even  when 


228 


MECHANICS   OF   ENGINEERING. 


tensile  or  compressive  stresses  are  also  present  on  the 
faces  considered. 

209.  Shearing  Stress  is  Always  of  the  Same  Intensity  on  the 
Four  Faces  of  an  Element. — (By  intensity  is  meant  per  unit 
of  area  ;  and  the  four  faces  referred  to  are  those  perpen- 
dicular to  the  paper  in  Fig.  208,  the  shearing  stress  being 
parallel  to  the  paper.) 

Let  dx  and  dz  be  the  width  and  height  of  the  element 
in  Fig.  208,  while  dy  is  its  thickness  perpendicular  to  the 
paper.  Let  the  intensity  of  the  shear  on  the  right  hand 
face  be  =ys,  that  on  the  top  face  =ps.  Then  for  the  ele- 
ment aw  a  free  body,  taking  moments  about  the  axis  0  per- 
pendicular to  paper,  we  have 

qs  dz  dy  X  dx — ps  dx  dy  x  dz  =0  .*.  gs  =ps 

(dx  and  dz  being  the  respective  lever  arms  of  the  forces 
qs  dz  dy  and  ps  dx  dy.) 

Even  if  there  were  also  tensions  (or  compressions)  on 
one  or  both  pairs  of  faces  their  moments  about  0  would 
balance  (or  fail  to  do  so  by  a  differential  of  a  higher  order) 
independently  of  the  shears,  and  the  above  result  would 
still  hold. 

210.  Table  of  Moduli  for  Shearing. 


if 

Ea 

8* 

8 

Material. 

i.e.  6  at  elastic 
limit. 

Mod.  of  Elasticity 
for  Shearing. 

(Elastic  limit.) 

(Rupture.) 
Ibs.  per  sq.  in. 

arc  in  ^-measure. 

Ibs.  per  sq.  in. 

Ibs.  per  sq.  in 

Soft  Steel, 

9,000,000 

70,000 

Hard  Steel, 

0.0032 

14,000,000 

45,000 

90,000 

Cast  Iron, 

0.0021 

7,000,000 

15,000 

30,000 

Wrought  Iron, 

0.0022 

9,000,000 

20,000 

50,000 

Brass, 

5,000,000 

Glass, 

Wood,  across! 
fibre,            1 

1,500 
to 
8,000 

Wood,  along  ( 
fibre,            1 

500 
to 
1,200 

SHEARING.  229 

As  in  the  tables  for  tension  and  compression,  the  above 
values  are  averages.  The  true  values  may  differ  from 
these  as  much  as  30  per  cent,  in  particular  cases,  accord" 
ing  to  the  quality  of  the.  specimen. 

211.  Punching  rivet  holes  in  plates  of  metal  requires  the 
overcoming  of  the  shearing  resistance  along  the  convex 
surface  of  the  cylinder  punched  out.     Hence  if  d  =  diam- 
eter of  hole,  and  t=  the  thickness  of  the  plate,  the  neces- 
sary force  for  the  punching,  the  surface  sheared  being 
F=  fad,  is 

P=Sfad        .        .        .        .      (2) 

Another  example  of  shearing  action  is  the  "  stripping  " 
of  the  threads  of  a  screw,  when  the  nut  is  forced  off  lon- 
gitudinally without  turning,  and  resembles  punching  in 
its  nature. 

212.  EandEs;  Theoretical  Relation. — In  case  a  rod  is  in 
tension  within  the  elastic  limit,  the  relative  (linear)  lateral 
contraction  (let  this  =m)  is  so  connected  with  Et  and  Em 
that  if  two  of  the  three  are  known  the  third  can  be  de- 
duced theoretically.     This  relation  is  proved  as  follows, 
by  Prof.  Burr.     Taking  an  elemental  cube  with  four  of  its 
faces  at  45°  with  the  axis  of  the  piece,  Fig.  209,  the  axial 
half-diagonal  AD  becomes  of  a  length  ADI=AD+£.AD 
under  stress,  while  the  transverse  half  diagonal  contracts 
to  a  length  B'D'=AD — m.AD.     The  angular  distortion  d 


E 

A 


PIG.  209.    §  212.  FIG.  210. 


230  MECHANICS   OF   ENGINEERING. 

is  supposed  very  small  compared  with  90°  and  is  due  to 
the  shear  ps  per  unit  of  area  on  the  face  BG  (or  BA\ 
From  the  figure  we  have 


tan(45°_)  =  ==l_^e,  appro*. 

[But,  Fig.  210,  tan(45°  —  x)=l  —  2#  nearly,  where  a;  is  a 
small  angle,  for,  taking  CA=  unity=  AE,  tan  AD=  AF= 
AE—EF.  Now  approximately  EF=  'EG.  ^/2  and  EG= 
~BI)^/2=x^/2  .'.  AF=  l—2x  nearly.]  Hence 

1  —  <5=  1  —  m  —  e  ;  or  S=  m+e        .         .       (2) 

Eq.  (2)  holds  good  whatever  the  stresses  producing  the 
deformation,  but  in  the  present  case  of  a  rod  in  tension, 
if  it  is  an  isotrope,  and  if  p  =  tension  per  unit  of  area  on 
its  transverse  section,  (see  §  182,  putting  «=45°),  we  have 
Et=p+s  and  fi*=(ps  on  BC)+3=y^p+d.  Putting  also 
(m  :  £)=r,  whence  ra=r£,  eq.  (2)  may  finally  be  written 


Prof.  Bauschinger,  experimenting  with  cast  iron  rods, 
found  that  in  tension  the  ratio  m  :  e  was  =  fib*  as  an  average, 
which  in  eq.  (3)  gives 


His  experiments  on  the  torsion  of  cast  iron  rods  gave 
#s=  6,000,000  to  7,000,000  Ibs.  per  sq.  inch.  By  (4),  then, 
Et  should  be  15,000,000  to  17,500,000  which  is  approxi- 
mately true  (§  203). 

Corresponding  results  may  be  obtained  for  short  blocks 
in  compression,  the  lateral  change  being  a  dilatation  in- 
stead of  a  contraction. 


SHEARING.  231 

£13.  Examples  in  Shearing.  —  EXAMPLE  1.  —  Required  the 
proper  length,  a,  Fig.  211,  to 
guard  against  the  shearing  off, 
along  the  grain,  of  the  portion 
ab,  of  a  wooden  tie-rod,  the  force 
P  being  =  2  tons,  and  the  width 
of  the  tie  =  4  inches.  Using  a 
value  of  S'  =  100  Ibs.  per  sq.  in., 
we  put  5a#'=  4,000  cos  45°;  i.e. 
FIG.  an.  a=  (4,000  xO.  707)  -f-  (4x100)=  7.07 

inches. 

EXAMPLE  2.  —  A  fy  in.  rivet  of  wrought  iron,  in  single 
shear  (see  Fig.  205)  has  an  ultimate  shearing  strength 
P=  FS  =  i^THftS^  i^X  }&  )2  x  50,000=  30,050  Ibs.  F  or  safety, 
putting  #'=8,000  instead  of  S,P'=  4,800  Ibs.  is  its  safe 
shearing  strength  in  single  shear. 

The  wrought  iron  plate,  to  be  secure  against  the  side- 
crushing  in  the  hole,  should  have  a  thickness  t,  computed 
thus  : 

P'=tdC';  or  4,800=  *.^xX2,000  .%  *=0.46  in. 


If  the  plate  were  only  0.23  in.  thick  the  safe  value  of  P 
would  be  only  ^  of  4,800. 

EXAMPLE  3.  —  Conversely,  given  a  lap-joint,  Fig.  205,  in 
which  the  plates  are  *£  in.  thick  and  the  tensile  force  on 
the  joint  =  600  Ibs.  per  linear  inch  of  seam,  how  closely 
must  y^  inch  rivets  be  spaced  in  one  row,  putting  #'=8,000 
and  C"=12,000  Ibs.  per  sq.  in.  ?  Let  the  distance  between 
centres  of  rivets  be  =x  (in  inches),  then  the  force  upon 
each  rivet  =600o;,  while  its  section  1^=0.44  sq.  in.  Having 
regard  to  the  shearing  strength  of  the  rivet  we  put  600x= 
0.44x8,000  and  obtain  #=5.86  in.;  but  considering  that  the 
safe  crushing  resistance  of  the  hole  is  =  i^^.  12,000= 
2,250  Ibs.,  600x=2,250  gives  z=3.75  inches,  which  is  the 
pitch  to  be  adopted.  What  is  the  tensile  strength  of  the 
reduced  sectional  area  of  the  plate,  with  this  pitch  ? 


17SS87 


232  MECHANICS   OF   ENGINEERING. 

EXAMPLE  4. — Double  butt-joint ;  (see  Fig.  207) ;  %  inch 
plate;  ^  in.  rivets;  T=  (7=12,000 ;  £'=8,333;  width  of 
plates =14  inches.  Will  one  row  of  rivets  be  sufficient  at 
each  side  of  joint,  if  #=30,000  Ibs.?  The  number  of  rivets 
=  ?  Here  each  rivet  is  in  double  shear  and  has  therefore 
a  double  strength  as  regards  shear.  In  double  shear  the 
safe  strength  of  each  rivet  =ZFS'= 7,333  Ibs.  Now  30,000 -r- 
7,333=4.0  (say).  With  the  four  rivets  in  one  row  the  re- 
duced sectional  area  of  the  main  plate  is  =[14 — 4x  ^]  X3/a 
=4.12  sq.  in.,  whose  safe  tensile  strength  is  =^T27'=4.12x 
12,000=49,440  Ibs.;  which  is  >  30,000  Ibs.  .-.  main  plate  is 
safe  in  this  respect.  But  as  to  side-crushing  in  holes 
in  main  plate  we  find  that  C't.d  (i.e.  12,000  X 3/8 X  %  =3,375 
Ibs.)  is  <^Q  i.e. <7,500  Ibs.,  the  actual  force  on  side  of 
hole.  Hence  four  rivets  in  one  row  are  too  few  unless 
thickness  of  main  plate  be  doubled.  Will  eight  in  one 
row  be  safe  ? 

213a.  (Addendum  to  §  206.)  Elasticity  of  Stone  and  Cements. 
— Experiments  by  Gen.  Gillmore  with  the  large  Watertown 
testing-machine  in  1883  resulted  as  follows  (see  p.  221  for 
notation) : 

With  cubes  of  Haverstraw  Freestone  (a  homogeneous  brown- 
stone)  from  1  in.  to  12  in.  on  the  edge,  E0  was  found  to  be 
from  900,000  to  1,000,000  Ibs.  per  sq.  in.  approximately  ;  and 
0  about  4,000  or  5,000  Ibs.  per  sq.  in.  Cubes  of  the  same 
range  of  sizes  of  Dyckerman's  Portland  cement  gave  E0  from 
1,350,000  to  1,630,000,  and  C  from  4,000  to  7,000,  Ibs.  per  sq. 
in.  Cubes  of  concrete  of  the  above  sizes,  made  with  the 
Newark  Cc.'s  Rosendale  cement,  gave  E0  about  538,000,  while 
cubes  of  cement-mortar,  and  some  of  concrete,  both  made  with 
National  Portland  cement,  showed  E0  from  800,000  to  2,000,- 
000  Ibs.  per  sq.  in. 

The  compressibility  of  brick  piers  12  in.  square  in  section 
and  16  in.  high  was  also  tested.  They  were  made  of  common 
North  River  brick  with  mortar  joints  f  in.  thick,  and  showed 
a  value  for  E0  of  about  300,000  or  400,000,  while  at  elastic 
limit  C"  was  on  the  average  1,000,  Ibs.  per  sq.  in. 


TORSION. 


233 


CHAPTER  II. 
TORSION. 

214.  Angle  of  Torsion  and  of  Helix.  When  a  cylindrical 
beam  or  shaft  is  subjected  to  a  twisting  or  torsional  action, 
i.  e.  when  it  is  the  means  of  holding  in  equilibrium  two 
couples  in  parallel  planes  and  of  equal  and  opposite  mo- 
ments, the  longitudinal  axis  of  symmetry  remains  straight 
(and  the  elements  along  it  exper- 
ience no  stress  (whence  it  may  be 
called  the  "line  of  no  twist"), 
while  the  lines  originally  parallel  to 
it  assume  the  form  of  helices,  each 
element  of  which  is  distorted  in  its  angles  (originally 
right  angles),  the  amount  of  distortion  being  assumed  pro- 
portional to  the  radius  of  the  helix.  The  directions  of  the 


faces  of  any  element  were  originally  as  follows  :  two  radial, 
two  in  consecutive  transverse  sections,  and  the  other  two 
tangent  to  two  consecutive  circular  cylinders  whose  com- 
mon axis  is  that  of  the  shaft.  E.g.  in  Fig.  212  we  have 
an  unstrained  shaft,  while  in  Fig.  213  it  holds  the  two 


234  MECHANICS   OF   ENGINEEKIXG. 

couples  (of  equal  moment  P  a  =  Q  b)  in  equilibrium.  These 
couples  act  in  parallel  planes  perpendicular  to  the  axis  of 
the  prism  and  a  distance,  lt  apart.  Assuming  that  the 
transverse  sections  remain  plane  and  parallel  during  tor- 
sion, any  surface  element,  m,  which  in  Fig.  212  was  entire- 
ly right-angled,  is  now  distorted.  Two  of  its  angles  have 
been  increased,  two  diminished,  by  an  amount  8,  the  angle 
between  the  helix  and  a  line  parallel  to  the  axis.  Suppos- 
ing m  to  be  the  most  distant  of  any  element  from  the  axis, 
this  distance  being  e,  any  other  element  at  a  distance  z 

from  the  axis  experiences  an  angular  distortion  =  -  d. 

6 

If  now  we  draw  0  B'  parallel  to  O'A  the  angle  B  0  B', 
=a,  is  called  the  Angle  of  Torsion,  while  d  may  be  called  the 
helix  angle;  the  former  lies  in  a  transverse  plane,  the  latter 
in  a  plane  tangent  to  the  cylinder.  Now 

tan  d  =  (linear  arc  B  B')-±-l ;  but  lin.  arc  B  B'=  ea\  hence, 
putting  8  for  tan  8,  (8  being  small) 

»-? d) 

(8  and  a  both  in  ic  measure). 

215.  Shearing  Stress  on  the  Elements.  The  angular  distor- 
tion, or  shearing  strain,  8,  of  any  element  (bounded  as  al- 
ready described)  is  due  to  the  shearing  stresses  exerted  on 
it  by  its  neighbors  on  the  four  faces  perpendicular  to  the 

tangent  plane  of  the  cylindri- 
cal shell  in  which  the  element 
is  situated.  Consider  these 
neighboring  elements  of  an 
outside  element  removed,  and 
the  stresses  put  in ;  the  latter 
are  accountable  for  the  dis- 
tortion of  the  element  and  so^ 


TORSION.  235 

hold  it  in  equilibrium.  Fig.  214  shows  this  element 
"free."  "Within  the  elastic  limit  d  is  known  to  be  propor- 
tional to  j9,,  the  shearing  stress  per  unit  of  area  on  the 
faces  whose  relative  angular  positions  have  been  changed. 
That  is,  from  eq.  (1)  §  208,  d=p^Es',  whence,  see  (1)  of 
§214, 


(2) 


In  (2)  pa  and  e  both  refer  to  a  surface  element,  e  being 
the  radius  of  the  cylinder,  and  p,.  the  greatest  intensity  of 
shearing  stress  existing  in  the  shaft.  Elements  lying  nearer 
the  axis  suffer  shearing  stresses  of  less  intensity  in  pro- 
portion to  their  radial  distances,  i.e.,  to  their  helix-angles. 
That  is,  the  shearing  stress  on  that  face  of  the  element 
which  forms  a  part  of  a  transverse  section  and  whose  dis- 
tance from  the  axis  is  z,  is  p}  =—  ps,  per  unit  of  area,  and 

the  total  shear  on  the  face  is  pdF,  dF  being  the  area  of  the 
face. 

216.  Torsional  Strength. — We  are  now  ready  to  expose  the 
full  transverse  section  of  a  shaft  under  torsion,  to  deduce 
formulae  of  practical  utility.  Making  a  right  section  of 
the  shaft  of  Fig.  213  anywhere  between  the  two  couples 
and  considering  the  left  hand  portion  as  a  free  body,  the 
forces  holding  it  in  equilibrium  are  the  two  forces  P  of 
the  left-hand  couple  and  an  infinite  number  of  shearing 
forces,  each  tangent  to  its  circle  of  radius  z,  on  the  cross 
section  exposed  by  the  removal  of  the  right-hand  portion. 
The  cross  section  is  assumed  to  remain  plane  during  tor- 
sion, and  is  composed -of  an  infinite  number  of  dF's,  each 
being  the  area  of  an  exposed  face  of  an  element ;  see  Fig. 
215. 


MECHANICS  OF  ENGINEEKING. 


Each  elementary  shearing  force  =  —  p*dF,  and  z  is  its 
lever  arm  about  the  axis  Oo .  For  equilibrium,  2*  (mom.) 
about  the  axis  Oo  must  =0 ;  i.e.  in  detail 


or,  reducing, 


P*  fz*dF=Pa;  or,  M. 
eJ  e 


(3) 


Eq.  (3)  relates  to  torsional  strength,  since  it  contains  ps,  the 
greatest  shearing  stress  induced  by  the  torsional  couple, 
whose  moment  Pa  is  called  the  Moment  of  Torsion,  the 
stresses  in  the  cross  section  forming  a  couple  of  equal  and 
opposite  moment. 

Jp  is  recognized  as  the  Polar  Moment  of  Inertia  of  the  cross 
section,  discussed  in  §  94  ;  e  is  the  radial  distance  of  the 
outermost  element,  and  =  the  radius  for  a  circular  shaft. 

217.  Torsional  Stiffness. — In  problems  involving  the  angle 
of  torsion,  or  deformation  of  the  shaft,  we  need  an  equa- 
tion connecting  Pa  and  a,  which  is  obtained  by  substitut- 
ing in  eq.  (3)  the  value  of  pB  in  eq.  (2),  whence 


Pa     . 


TORSION.  237 

From  this  it  appears  that  the  angle  of  torsion,  a,  is  propor- 
tional to  the  moment  of  torsion,  Pa,  within  the  elastic 
limit  ;  a  must  be  expressed  in  ^-measure.  Trautwine  cites  1° 
(i.e.  a=  0.0174)  as  a  maximum  allowable  value  for  shafts. 

218.  Torsional  Resilience  is  the  work  done  in  twisting  a 
shaft  from  an  unstrained  state  until  the  elastic  limit  is 
reached  in  the  outermost  elements.  If  in  Fig.  213  we 
imagine  the  right-hand  extremity  to  be  fixed,  while  the 
other  end  is  gradually  twisted  through  an  angle  a\  each 
force  J^*  of  the  couple  must  be  made  to  increase  gradually 
from  a  zero  value  up  to  the  value  Pl}  corresponding  to  at. 
In  this  motion  each  end  of  the  arm  a  describes  a  space 
=  K««i,  and  the  mean  value  of  the  force  =  ^4Pl  (compare 
§  196).  Hence  the  work  done  in  twisting  is 


.         .    (5) 
By  the  aid  of  preceding  equations,  (5)  can  be  written 

or  =  &L  (6) 

1 


If  for  ps  we  write  S'  (Modulus  of  safe  shearing)  we  have 
for  the  safe  resilience  of  the  shaft 


If  the  torsional  elasticity  of  an  originally  unstrained  shaft 
is  to  be  the  means  of  arresting  the  motion  of  a  moving 
mass  whose  weight  is  G,  (large  compared  with  the  parts 
intervening)  and  velocity  =v,  we  write  (§  133) 

V=£.ti 

<7    2' 

as  the  condition  that  the  shaft  shall  not  be  injured. 


238  MECHANICS   OF   ENGINEERING. 

219.  Polar  Moment  of  Inertia.  —  For  a  shaft  of  circular 
cross  section  (see  §  94)  7p=i^^r4;  for  a  hollow  cylinder 
I^y^rt—rf)  ;  while  for  a  square  shaft  7p=i^64,  b  being 
the  side  of  the  square  ;  for  a  rectangular  cross-section 
sides  b  and  h,  Ip=^bh(b2-\-h2).  For  a  cylinder  e=r;  if  hol- 
low, e=r  ,  the  greater  radius.  For  a  square,  6 


220.  Non-Circular  Shafts.  —  If  the  cross-section  is  not  cir- 
cular it  becomes  warped,  in  torsion,  instead  of  remaining 
plane.  Hence  the  foregoing  theory  does  not  strictly  ap- 
ply. The  celebrated  investigations  of  St.  Venant,  how- 
ever, cover  many  of  these  cases.  (See  §  708  of  Thompson 
and  Tait's  Natural  Philosophy  ;  also,  Prof.  Burr's  Elas- 
ticity and  Strength  of  the  Materials  of  Engineering).  His 
-results  give  for  a  square  shaft  (instead  of  the 


.         .        *?      .    (1) 

and  Pa^/jPp.,  instead  of  eq.  (3)  of  §  216,  pn  being  the 
greatest  shearing  stress. 

The  elements  under  greatest  shearing  strain  are  found 
at  the  middles  of  the  sides,  instead  of  at  the  corners,  when 
the  prism  is  of  square  or  rectangular  cross-section.  The 
warping  of  the  cross-section  in  such  a  case  is  easily  veri- 
fied by  the  student  by  twisting  a  bar  of  india-rubber  in 
his  fingers. 

221.  Transmission  of  Power.  —  Fig.  216.  Suppose  the  cog- 
wheel B  to  cause  A,  on  the 
same  shaft,  to  revolve  uni- 
formly and  overcome  a  resis- 
tance Q,  the  pressure  of  the 
teeth  of  another  cog-wheel, 
B  being  driven  by  still  another 
FIG.  216.  wheel.  The  shaft  AB  is  un- 


TORSION.  239 

der  torsion,  the  moment  of  torsion  being  =Pa=  Qb.  (Pl 
and  $1  the  bearing  reactions  have  no  moment  about  the 
axis  of  the  shaft).  If  the  shaft  makes  u  revolutions  per 
unit-time,  the  work  transmitted  (transmitted  ;  not  expend- 
ed in  twisting  the  shaft  whose  angle  of  torsion  remains 
constant,  corresponding  to  Pa)  per  unit-time,  i.e.  the  Power, 
is 

L=-P.2xa.u=27tuPa        .        .        .    (8) 

To  reduce  L  to  Horse  Power  (§  132),  we  divide  by  Nt 
the  number  of  units  of  work  per  unit-time  constituting 
one  H.  P.  in  the  system  of  units  employed,  i.e., 

Horse  Power  =H.  P.= 


For  example  ^=33,000  ft.-lbs.  per  minute,  or  =396,000 
inch  -Ibs.  per  minute  ;  or  =  550  ft.-lbs.  per  second.  Usually 
the  rate  of  rotation  of  a  shaft  is  given  in  revolutions  per 
minute, 

But  ei.  (8)  happens  to  contain  Pa  the  moment  of  torsion 
acting  to  maintain  the  constant  value  of  the  angle  of  tor- 
sion, and  since  for  safety  (see  eq.  (3)  §  216)  Par=S'Ip+e, 
with  Jp=  y^Ttr^  and  e=r  for  a  solid  circular  shaft,  we  have 
for  such  a  shaft 


(9) 


which  is  the  safe  H.  P.,  which  the  given  shaft  can  trans- 
mit at  the  given  speed.  /S"  may  be  made  7,000  Ibs.  per  sq. 
inch  for  wrought  iron  ;  10,000  for  steel,  and  5,000  for  cast- 
iron.  If  the  value  of  Pa  fluctuates  periodically,  as  when 
a  shaft  is  driven  by  a  connecting  rod  and  crank,  for  (H.  P.) 
we  put  wX(H.  P.),  m  being  the  ratio  of  the  maximum  to 
the  mean  torsional  moment  ;  m  =  about  1V2  under  ordi- 
nary circumstances  (Cotterill). 


240 


MECHANICS   OF   ENGINEERING. 


222.  Autographic  Testing  Machine. — The  principle  of  Pro£ 
Thurston's  invention  bearing  this  name  is  shown  in  Fig 


FIG.  217. 

217.  The  test-piece  is  of  a  standard  shape  and  size,  its 
central  cylinder  being  subjected  to  torsion.  A  jaw,  carry- 
ing a  handle  (or  gear-wheel  turned  by  a  worm)  and  a  drum 
on  which  paper  is  wrapped,  takes  a  firm  hold  of  one  end 
of  the  test-piece,  whose  further  end  lies  in  another  jaw 
rigidly  connected  with  a  heavy  pendulum  carrying  a  pen- 
cil free  to  move  axially.  By  a  continuous  slow  motion  of 
the  handle  the  pendulum  is  gradually  deviated  more  and 
more  from  the  vertical,  through  the  intervention  of  the 
test-piece,  which  is  thus  subjected  to  an  increasing  tor- 
sional  moment.  The  axis  of  the  test-piece  lies  in  the  axis 
of  motion.  This  motion  of  the  pendulum  by  means  of  a 
properly  curved  guide,  WR,  causes  an  axial  (i.e.,  parallel 
to  axis  of  test-piece)  motion  of  the  pencil  A,  as  well  as  an 
angular  deviation  /9  equal  to  that  of  the  pendulum,  and 
this  axial  distance  CF,=sT,  of  the  pencil  from  its  initial 
position  measures  the  moment  of  torsion=f>a=jPc  sin  )tf. 
As  the  piece  twists,  the  drum  and  paper  move  relatively 
to  the  pencil  through  an  angle  sUo  equal  to  the  angle 


TORSION. 


241 


of  torsion  «  so  far  attained.  The  abscissa  so  and  ordinate 
sT  of  the  curve  thus  marked  on  the  paper,  measure, 
when  the  paper  is  unrolled,  the  values  of  a  and  Pa  through 
all  the  stages  of  the  torsion.  Fig.  218  shows  typical 


ANG. TORSION       60 


iurves  thus  obtained.  Many  valuable  indications  are 
given  by  these  strain  diagrams  as  to  homogeneousness  of 
composition,  ductility,  etc.,  etc.  On  relaxing  the  strain 
at  any  stage  within  the  elastic  limit,  the  pencil  retraces 
its  path  ;  but  if  beyond  that  limit,  a  new  path  is  taken 
called  an  "elasticity  -line,"  in  general  parallel  to  the  first 
part  of  the  line,  and  showing  the  amount  of  angular  re- 
covery, BC,  and  the  permanent  angular  set,  OB. 

223.  Examples  in  Torsion.  —  The  modulus  of  safe  shearing 
strengtn,  S',  as  given  in  §  221,  is  expressed  in  pounds  per 
square  inch  ;  hence  these  two  units  should  be  adopted 
throughout  in  any  numerical  examples  where  one  of  the 
above  values  for  S'  is  used.  The  same  statement  applies 
to  the  modulus  of  shearing  elasticity,  .Z7S,  in  the  table  of 
§  210. 

EXAMPLE  1.—  Fig.  216.  With  P  =  1  ton,  a  =  3  ft.,  I  = 
10  ft.,  and  the  radius  of  the  cylindrical  shaft  r=2.5  inches, 
required  the  max.  shearing  stress  per  sq.  inch,  pa  the 
shaft  being  of  wrought  iron.  From  eq.  (3)  §  216 


which  is  a  safe  value  for  any  ferrous  metal. 


242 


MECHANICS  OF   ENGINEERING. 


EXAMPLE  2. — What  H.  P.  is  the  shaft  in  Ex.  1  transmit- 
ting, if  it  makes  50  revolutions  per  minute  ?  Let  u  = 
number  of  revolutions  per  unit  of  time,  and  N=  the  num- 
ber of  units  of  work  per  unit  of  time  constituting  one 
horse-power.  Then  H.  I*.=Pu%7ra-r-N,  which  for  the  foot- 
pound-minute system  of  units  gives 

H.  P.=2,000x50x2^x3--33,000=57^:  H.  P. 

EXAMPLE  3. — What  different  radius  should  be  given  to 
the  shaft  in  Ex.  1,  if  two  radii  at  its  extremities,  originally 
parallel,  are  to  make  an  angle  of  2°  when  the  given  moment 
of  torsion  is  acting,  the  strains  in  the  shaft  remaining  con- 
stant. From  eq.  (4)  §  217,  and  the  table  210,  with  a=^= 
0.035  radians  (i.e.  TT -measure),  and  .Z^y^rr4,  we  have 


2,000x36x120       _ 
^0.035x9,000,000 


.•.  r=2.04  inches. 


(This  would  bring  about  a  different  p.,  but  still  safe.)  The 
foregoing  is  an  example  in  stiffness. 

EXAMPLE  4. — A  working  shaft  of  steel  (solid)  is  to  trans- 
mit 4,000  H.  P.  and  make  60  rev.  per  minute,  the  maximum 
twisting  moment  being  1^  times  the  average;  required 
its  diameter.  d=14.74  inches.  Ans. 

EXAMPLE  5. — In  example  1,  pA=  2,930  Ibs.  per  square 
inch  ;  what  tensile  stress  does  this  imply  on  a  plane  at  45° 
with  the  pair  of  planes  on  which  ps  acts  ?  Fig.  219  shows 


p.dx* 


p.dx1 


FIG.  220. 


TORSION.  243 

a  small  cube,  of  edge  =dx,  (taken  from  the  outer  helix  of 
Fig.  215,)  free  and  in  equilibrium,  the  plane  of  the  paper 
being  tangent  to  the  cylinder  ;  while  220  shows  the  portion 
BDC,  also  free,  with  the  unknown  total  tensile  stresspcfcc2^ 
acting  on  the  newly  exposed  rectangle  of  area  =dxXdx^"2, 
p  being  the  unknown  stress  per  unit  of  area.  From  sym- 
metry the  stress  on  this  diagonal  plane  has  no  shearing 
component.  Putting  2  [components  normal  to  BD]=Q, 
we  have 

pdxz*/2=2dx2pscos4:50=dxtps^2~.'.p=ps    .      (1) 

That  is,  a  normal  tensile  stress  exists  in  the  diagonal 
plane  BD  of  the  cubical  element  equal  in  intensity  to  the 
shearing  stress  on  one  of  the  faces,  i.e.,  =2,930  Ibs.  per  sq. 
in.  in  this  case. 

Similarly  in  the  plane  AC  will  be  found  a  compressive 
stress  of  2,930  Ibs.  per  sq.  in.  If  a  plane  surface  had  been 
exposed  making  any  other  angle  than  45°  with  the  face  of 
the  cube  in  Fig.  219,  we  should  have  found  shearing  and 
normal  stresses  each  less  than  pa  per  sq.  inch.  Hence  the 
interior  dotted  cube  in  219,  if  shown  "  free  "  is  in  tension 
in  one  direction,  in  compression  in  the  other,  and  with 
no  shear,  these  normal  stresses  having  equal  intensities. 
Since  S'  is  usually  less  than  T'  or  C',  it  ps  is  made  =  >S" 
the  tensile  and  compressive  actions  are  not  injurious.  It 
follows  therefore  that  when  a  cylinder  is  in  torsion  any 
helix  at  an  angle  of  45°  with  the  axis  is  a  line  of  tensile, 
or  of  compressive  stress,  according  as  it  is  a  right  or  left 
handed  helix,  or  vice  versa. 

EXAMPLE  6. — A  solid  and  a  hollow  cylindrical  shaft,  of 
equal  length,  contain  the  same  amount  of  the  same  kind 
of  metal,  the  solid  one  fitting  the  hollow  of  the  other. 

Compare  their  torsional  strengths,  used  separately. 
The  solid  shaft  has  only  ^  the  strength  of  the  hollow 
one.  Ans. 


244  MECHANICS  OF 


CHAPTER  III. 

FLEXURE  OF  HOMOGENEOUS  PRISMS  UNDER 
PERPENDICULAR  FORCES  IN  ONE  PLANE. 

224.  Assumptions  of  the  Common  Theory  of  Flexure. — When 
a  prism  is  bent,  under  the  action  of  external  forces  per- 
pendicular to  it  and  in  the  same  plane  with  each  other,  it 
may  be  assumed  that  the  longitudinal  fibres  are  in  tension 
on  the  convex  side,  in  compression  on  the  concave  side, 
and  that  the  relative  stretching  or  contraction  of  the  ele- 
ments is  proportional  to  their  distances  from  a  plane  in- 
termediate between,  with  the  understanding  that  the  flex- 
ure is  slight  and  that  the  elastic  limit  is  not  passed  in  any 
element. 

This  "  common  theory  "  is  sufficiently  exact  for  ordinary 
engineering  purposes  if  the  constants  employed  are  prop- 
erly determined  by  a  wide  range  of  experiments,  and  in- 
volves certain  assumptions  of  as  simple  a  nature  as  possi- 
ble, consistently  with  practical  facts.  These  assumptions 
are  as  follows,  (for  prisms,  and  for  solids  with  variable  cross 
sections,  when  the  cross  sections  are  similarly  situated  as 
regards  a  central  straight  axis)  and  are  approximately 
borne  out  by  experiment : 

(1.)  The  external  or  "  applied  "  forces  are  all  perpendicu- 
lar to  the  axis  of  the  piece  and  lie  in  one  plane,  which  may 
be  called  the  force-plane;  the  force-plane  contains  the 
axis  of  the  piece  and  cuts  each  cross-section  symmetri- 
cally; 

(2.)  The  cross-sections  remain  plane  surfaces  during 
flexure  ; 

(3.)  There  is  a  surface  (or,  rather,  sheet  of  elements) 
which  is  parallel  to  the  axis  and  perpendicular  to  the 
force-plane,  and  along  which  the  elements  of  the  solid  ex- 


FLEXURE. 


245 


perieuce  no  tension  nor  compression  in  an  axial  direction, 
this  being  called  the  Neutral  Surface ; 

(4.)  The  projection  of  the  neutral  surface  upon  the  force 
plane  (or  a  ||  plane)  being  called  the  Neutral  Line  or  Elastic 
Curve,  the  bending  or  flexure  of  the  piece  is  so  slight  that 
an  elementary  division,  ds,  of  the  neutral  line  may  be  put 
=dx,  its  projection  on  a  line  parallel  to  the  direction  of 
the  axis  before  flexure  ; 

(5.)  The  elements  of  the  body  contained  between  any 
two  consecutive  cross-sections,  whose  intersections  with 
the  neutral  surface  are  the  respective  Neutral  Axes  of  the 
sections,  experience  elongations  (or  contractions,  accord- 
ing as  they  are  situated  on  one  side  or  the  other  of  the 
neutral  surface),  in  an  axial  direction,  whose  amounts  are 
proportional  to  their  distances  from  the  neutral  axis,  and 
indicate  corresponding  tensile  or  compressive  stresses ; 

(6.)  Et=Ec-, 

(7.)  The  dimensions  of  the  cross-section  are  small  com- 
pared with  the  length  of  the  piece  ; 

(8.)  There  is  no  shear  perpendicular  to  the  force  plane 
on  internal  surfaces  perpendicular  to  that  plane. 

In  the  locality  where  any  one  of  the  external  forces  is 
Applied,  local  stresses  are  of  course  induced  which  demand 
separate  treatment.  These  are  not  considered  at  present. 

225.  Illustration. — Consider  the  case  of  flexure  shown  in 
Fig.  221.  The  external  forces  are  three  (neglecting  the 


246 


MECHANICS   OF   ENGINEERING. 


weight  of  the  beam),  viz.:  Plt  P2,  and  P3.     Pl  and  P3  are 
loads,  P2  the  reaction  of  the  support. 

The  force  plane  is  vertical.  N^L  is  the  neutral  line  or 
elastic  curve.  NA  is  the  neutral  axis  of  the  cross -section 
at  m ;  this  cross-section,  originally  perpendicular  to  the 
sides  of  the  prism,  is  during  flexure  ~|  to  their  tangent 
planes  drawn  at  the  intersection  lines  ;  in  other  words,  the 
side  view  QNB,  of  any  cross-section  is  perpendicular  to 
the  neutral  line.  In  considering  the  whole  prism  free  we 
have  the  system  Plt  P2,  and  P3  in  equilibrium,  whence 
from  2T=0  we  have  P2=Pi+P3,  and  from  I*  (mom.  about 
O)  =  0,  P3Z3=  PI?!-  Hence  given  Pl  we  may  determine  the 
other  two  external  forces.  A  reaction  such  as  P2  is  some- 
times called  a  supporting  force.  The  elements  above  the 
neutral  surface  N^LS&re  in  tension ;  those  below  in  com- 
pression (in  an  axial  direction). 

226.  The  Elastic  Forces. — Conceive  the  beam  in  Fig.  221 
separated  into  two  parts  by  any  transverse  section  such 
as  QA,  and  the  portion  NiON,  considered  as  a  free  body 
in  Fig.  222.  Of  this  free  body  the  surface  QAB  is  one  of 


PLEXURfc.  247 

the  bounding  surfaces,  but  was  originally  an  internal  sur- 
face of  the  beam  m  Fig.  221.  Hence  in  Fig.  222  we  must 
put  in  the  stresses  acting  on  all  the  dF's  or  elements  of  area 
of  QAB.  These  stresses  represent  the  actions  of  the  body 
taken  away  upon  the  body  which  is  left,  and  according  to 
assumptions  (5),  (6)  and  (8)  consist  of  normal  stresses  (ten- 
sion or  compression)  proportional  per  unit  of  area,  to  the 
distance,  2,  of  the  cLF's  from  the  neutral  axis,  and  of  shear- 
ing stresses  parallel  to  the  force-plane  (which  in  most 
cases  will  be  vertical). 

The  intensity  of  this  shearing  stress  on  any  dF  varies 
with  the  position  of  the  dF  with  respect  to  the  neutral 
axis,  but  the  law  of  its  variation  will  be  investigated  later 
(§§  253  and  254).  These  stresses,  called  the  Elastic  Forces 
of  the  cross-section  exposed,  and  the  external  forces  Pl  and 
P2,  form  a  system  in  equilibrium.  We  may  therefore  ap- 
ply any  of  the  conditions  of  equilibrium  proved  in  §  38. 


227.  The  Neutral  Axis  Contains  the  Centre  of  Gravity  of  the 
Cross-Section.  —  Fig.  222.  Let  e=  the  distance  of  the  outer- 
most element  of  the  cross-section  from  the  neutral  axis,  and 
the  normal  stress  per  unit  of  area  upon  it  be  =p,  whether 
tension  or  compression.  Then  by  assumptions  (5)  and  (6), 
§  224,  the  intensity  of  normal  stress  on  any  dF  is  =  -1  p 
and  the  actual 


normal  stress  on  any  c^is^  ±-  pdF        .        (1) 

This  equation  is  true  for  dF'a  having  negative  «'s,  i.e. 
on  the  other  side  of  the  neutral  axis,  the  negative  value 
of  the  force  indicating  normal  stress  of  the  opposite  char- 
acter ;  for  if  the  relative  elongation  (or  contraction)  of  two 
axial  fibres  is  the  same  for  equal  z's,  one  above,  the  other 
below,  the  neutral  surface,  the  stresses  producing  the 
changes  in  length  are  also  the  same,  provided  E^—EC\  see  §§ 
184  and  201. 


248  MECHANICS   OF   ENGINEERING. 

For  this  free  body  in  equilibrium  put  IX=Q  (Xisa 
horizontal  axis).  Put  the  normal  stresses  equal  to  their 
X  components,  the  flexure  being  so  slight,  and  the  X  com- 
ponent of  the  shears  =  0  for  the  same  reason.  This  gives 
(see  eq.  (1)  ) 

C-  pdF=  0  ;  i.e.  IL    CdFz=  0  ;  or,  £  Fl=0        (2) 
«/    e  e  J  e 

IH  which  2=  distance  of  the  centre  of  gravity  of  the  cross- 
section  from  the  neutral  axis,  from  which,  though  un- 
known in  position,  the  a's  have  been  measured  (see  eq. 
(4)  §23). 

In  eq.  (2)  neither  p^-e  nor  F  can  be  zero  .-.  z  'must  =  0 ; 
i.e.  the  neutral  axis  contains  the  centre  of  gravity.  Q.  E.  D. 
[If  the  external  forces  were  not  all  perpendicular  to  the 
beam  this  result  would  not  be  obtained,  necessarily.] 

228.  The  Shear. — The     "  total    shear,"    or    simply    the 
"  shear,"  in  the  cross-section  is  the  sum   of  th.e  vertical 
shearing  stresses  on  the  respective  dF's.     Call  this  sum 
J,  and  we  shall  have  from  the  free  body  in  Fig.  222,  by 
putting  2T=0  (Y being  vertical) 

P2—Pl—J=Q.'.J=P2—Pl        .        .         (3) 

That  is,  the  shear  equals  the  algebraic  sum  of  the  ex- 
ternal forces  acting  on  one  side  (only)  of  the  section  con- 
sidered. This  result  implies  nothing  concerning  its  mode 
of  distribution  over  the  section. 

229.  The  Moment.— By   the   "Moment    of    Flexure"  or 
simply  the  Moment,  at  any  cross- section  is  meant  the  sum 
of  the  moments  of  the  elastic  forces  of  the  section,  taking 
the  neutral  axis  as  an  axis  of  moments.     In  this  summa- 
tion the  normal  stresses  appear  alone,  the  shear  taking  no 
part,  having  no  lever  arm  about  the  neutral  axis.     Hence, 
Fig.  222,  the  moment  of  flexure 


1028 

FLEXURE.  249 


This  function,     CdFz2,  of  the  cross-section  or  plane  figure 

is  the  quantity  called  Moment  of  Inertia  of  a  plane  figure, 
§  85.  For  the  free  body  in  Fig.  222,  by  putting  ^(mom.s 
about  the  neutral  axis  NA)=0,  we  have  then 


=0,  or  in  general,    ^=M    .   (5) 


in  which  M  signifies  the  sum  of  moments,  about  the  neutral 
axis  of  the  section,  of  all  the  forces  acting  on  the  free  body 
considered,  exclusive  of  the  elastic  forces  of  the  exposed 
section  itself. 


230.  Strength  in  Flexure. — Eq.  (5)  is  available  for  solving 
problems  involving  the  Strength  of  beams  and  girders,  since 
it  contains  p,  the  greatest  normal  stress  per  unit  of  area  to 
be  found  in  the  section. 

In  the  cases  of  the  present  chapter,  where  all  the  exter- 
nal forces  are  perpendicular  to  the  prism  or  beam,  and 
have  therefore  no  components  parallel  to  the  beam,  i.e.  to 
the  axis  X,  it  is  evident  that  the  normal  stresses  in  any 
section,  as  QB  Fig.  222,  are  equivalent  to  a  couple  ;  for  the 
condition  ^Y=0  falls  entirely  upon  them  and  cannot  be 
true  unless  the  resultant  of  the  tensions  is  equal,  parallel, 
and  opposite  to  that  of  the  compressions.  These  two  equal 
and  parallel  resultants,  not  being  in  the  same  line,  form  a 
couple  (§  28),  which  we  may  call  the  stress-couple.  The 
moment  of  this  couple  is  the  "  moment  of  flexure  "  p—  ,  and 
it  is  further  evident  that  the  remaining  forces  in  Fig.  222, 
viz.:  the  shear  J  and  the  external  forces  Pl  and  P2,  are 
equivalent  to  a  couple  of  equal  and  opposite  moment  to 
the  one  formed  by  the  normal  stresses. 


M.T. 


250 


MECHANICS   OF  ENGINEERING. 


231,  Flexural  Stiffness. — The  neutral  line,  or  elastic  curve* 
containing  the  centres  of  gravity  of  all  the  sections,  was 
originally  straight ;  its  radius  of  curvature  at  any  point, 
as  N,  Fig.  222,  during  flexure  may  be  introduced  as  fol- 
lows. QB  and  U'V  are  two  consecutive  cross-sections, 
originally  parallel,  but  now  inclined  so  that  the  intersec- 
tion (7,  found  by  prolonging  them  sufficiently,  is  the  centre 
of  curvature  of  the  ds  (put  =dx)  which  separates  them  at 
N,  and  CG=p=  the  radius  of  curvature  of  the  elastic 
curve  at  N.  From  the  similar  triangles  U'UGf&nd  GNCwe 
have  dX  :  dx  : :  e :.  p,  in  which  dX  is  the  elongation,  U'  U,  of  a 
portion,  originally  =dx,  of  the  outer  fibre.  But  the  rela- 
tive elongation  e=  -=—  of  the  latter  is,  by  §184,  within  the 


elastic  limit,  =-^L.*.  —  =—  and  eq.  (5)  becomes 
E     E     p 


(6) 


From  (6)  the  radius  of  curvature  can  be  computed.  E= 
the  value  of  Et=Ec,  as  ascertained  from  experiments  in 
bending. 

To  obtain  a  differential  equation  of  the  elastic  curve,  (6) 
may  be  transformed  thus,  Fig.  223.  The  curve  being  very 

AXI8X  flat,    consider    two    consecutive 

ds's  with  equal  dx's ;  they  may 
be  put  =  their  dx's.  Produce 
the  first  to  intersect  the  dy  of  the 
second,  thus  cutting  off  the  d*y, 
i.e.  the  difference  between  two 
consecutive  cfa/'s.  Drawing  a  per- 
pendicular to  each  ds  at  its  left 
extremity,  the  centre  of  curva- 
ture G  is  determined  by  their  in- 
tersection, and  thus  the  radius 
of  curvature  p.  The  two  shaded 
triangles  have  their  small  angles 


FLEXURE.  251 

equal,  and  d?y  is  nearly  perpendicular  to  the  prolonged 
ds  ;  hence,  considering  them  similar,  we  have 


and  hence  from  eq.  (6)  we  have 

(approx.)     ±ElQl=M    .        .         (7) 

as  a  differential  equation  of  the  elastic  curve.  From  this 
the  equation  of  the  elastic  curve  may  be  found,  the  de- 
flections at  different  points  computed,  and  an  idea  thus 
formed  of  the  stiffness.  All  beams  in  the  present  chap- 
ter being  prismatic  and  homogeneous  both  E  and  /  are  the 
same  (i.e.  constant)  at  all  points  of  the  elastic  curve.  In 
using  (7)  the  axis  JTmust  be  taken  parallel  to  the  length 
of  the  beam  before  flexure,  which  must  be  slight  ;  the 
minus  sign  in  (7)  provides  for  the  case  when  d-y+dy?  is  es- 
sentially negative. 

232.  Resilience  of  Flexure.  —  If  the  external  forces  are  made 
to  increase  gradually  from  zero  up  to  certain  maximum 
values,  some  of  them  may  do  work,  by  reason  of  their 
points  of  application  moving  through  certain  distances 
due  to  the  yielding,  or  flexure,  of  the  body.  If  at  the  be- 
ginning and  also  at  the  end  of  this  operation  the  body  is 
at  rest,  this  work  has  been  expended  on  the  elastic  resis- 
tance of  the  body,  and  an  equal  amount,  called  the  work 
of  resilience  (or  springing-back),  will  be  restored  by  the 
elasticity  of  the  body,  if  released  from  the  external  forces, 
provided  the  elastic  limit  has  not  been  passed.  The  energy 
thus  temporarily  stored  is  of  the  potential  kind  ;  see  §§ 
148,  180,  196  and  218. 

232a.  Distinction  Between  Simple,  and  Continuous,  Beams  (or 
"Girders").  —  The  external  forces  acting  on  a  beam  consist 


252 


MECHANICS   OF   ENGINE  EIIING. 


generally  of  the  loads  and  the  "  reactions  "  of  the  sup- 
ports. If  the  beam  is  horizontal  and  rests  on  two  supports 
only,  the  reactions  of  those  supports  are  easily  found  by 
elementary  statics  [§  36]  alone,  without  calling  into  ac- 
count the  theory  of  flexure,  and  the  beam  is  said  to  be  a 
Simple  Beam,  or  girder ;  whereas  if  it  is  in  contact  with 
more  than  two  supports,  being  "  continuous,"  therefore, 
over  some  of  them,  it  is  a  Continuous  Girder  (§  271).  The 
remainder  of  this  chapter  will  deal  only  with  simple 
beams. 


ELASTIC  CURVES. 

233.  Case  I.  Horizontal  Prismatic  Beam,  [Supported  at  Both 
Ends,  With  a  Central  Load,  Weight  of  Beam  Neglected. — Fig. 
224.  First  considering  the  whole  beam  free,  we  find  each 


(. m 


IB 


FIG.  224.    §  233. 


reaction  to  be  =  >£P.  AOB  is  the  neutral  line  ;  required 
the  equation  of  the  portion  OB  referred  to  0  as  an  origin, 
and  to  the  tangent  line  through  0  as  the  axis  of  X.  To 
do  this  consider  as  free  the  portion  mB  between  any  sec- 
tion m  on  the  right  of  0  and  the  near  support,  in  Fig. 
225.  The  forces  holding  this  free  body  in  equilibrium 


FIG.  225. 


ELASTIC   CURVES.  253 


are  the  one  external  force  ]^P,  and  the  elastic  forces  act- 
ing on  the  exposed  surface.  The  latter  consist  of  «7,  the 
shear,  and  the  tensions  and  compressions  represented  in 
the  figure  by  their  equivalent  "  stress-couple."  Selecting 
N,  the  neutral  axis  of  TO,  as  an  axis  of  moments  (that  J 
may  not  appear  in  the  moment  equation)  and  putting 
Jf  (mom)  =0  we  have 


Fig.  226  shows  the  elastic  curve  OB  in  its  purely  geomet- 
rical aspect,  much  exaggerated.  For  axes  and  origin  as  in 
figure  d^y-^-dx2  is  positive. 

Eq.  (1)  gives  the  second  x-derivative  of  y  equal  to  a 
function  of  x.  Hence  the  first  x-derivative  of  y  will  be 
equal  to  the  oj-anti-derivative  of  that  function,  plus  a  con- 
stant, C.  (By  anti-derivative  is  meant  the  converse  of  de- 
rivative, sometimes  called  integral  though  not  in  the  sense 
of  summation).  Hence  from  (1)  we  have  (El  being  a  con- 
stant factor  remaining  undisturbed) 


(2)'  is  an  equation  between  two  variables  dy-±- dx  and  x,  and 
holds  good  for  any  point  between  0  and  B\  dy-r-dx  de- 
noting the  tang,  of  a,  the  slope,  or  angle  between  the  tan- 
gent line  and  X.  At  0  the  slope  is  zero,  and  x  also  zero ; 
hence  at  0  (2)'  becomes 

#7x0=0— 0+  C 

which  enables  us  to  determine  the  constant  C,  whose  value 
must  be  the  same  at  0  as  for  all  points  of  the  curve. 
Hence  (7=0  and  (2)'  becomes 


254  MECHANICS  OF  ENGINEERING. 


dx 


from  which  the  slope,  tan.  a,  (or  simply  a,  in  /r-measure  ; 
since  the  angle  is  small)  may  be  found  at  any  point.  Thus 
at  B  we  have  x=y2l  and  dy-s-dx=al,  and 


_ 

''"'"IB  '  El 

Again,  taking  the  aj-anti-derivative  of  both  members  of  eq. 
(2)  we  have 


and  since  at  0  both  x  and  y  are  zero,  0'  is  zero.     Hence 
the  equation  of  the  elastic  curve  OB  is 


To  compute  the  deflection  of  0  from  the  right  line  join- 
ing A  and  B  in  Fig.  224,  i.e.  BK,  =d,  we  put  x=tfl  in  (3),  y 
being  then  =rf,  and  obtain 


Eq.  (3)  does  not  admit  of  negative  values  for  x  ;  for  if 
the  free  body  of  Fig.  225  extended  to  the  left  of  0,  the  ex. 
ternal  forces  acting  would  be  P,  downward,  at  0  ;  and  }4P, 
upward,  at  B,  instead  of  the  latter  alone  ;  thus  altering 
the  form  of  eq.  (1).  From  symmetry,  however,  we  know 
that  the  curve  AO,  Fig.  224,  is  symmetrical  with  OB  about 
the  vertical  through  Q. 


ELASTIC   CURVES.  255 

233a.  Load  Suddenly  Applied.  —  Eq.  (4)  gives  the  deflection 
d  corresponding  to  the  force  or  pressure  P  applied  at  the 
middle  of  the  beam,  and  is  seen  to  be  proportional  to  it. 
If  a  load  G  hangs  at  rest  from  the  middle  of  the  beam, 
P=  G  ;  but  if  the  load  G,  being  initially  placed  at  rest 
upon  the  unbent  beam,  is  suddenly  released  'from  the  ex- 
ternal constraint  necessary  to  hold  it  there,  it  sinks  and 
deflects  the  beam,  the  pressure  P  actually  felt  by  the.beam 
varying  with  the  deflection  as  the  load  sinks.  What  is 
the  ultimate  deflection  dm  ?  Let  Pm=  the  pressure  be- 
tween the  load  and  the  beam  at  the  instant  of  maximum 
deflection.  The  work  so  far  done  in  bending  the  beam 
=  i^Pmc?m.  The  potential  energy  given  up  by  the  load 
«=  Gdm,  while  the  initial  and  final  kinetic  energies  are  both 
nothing. 

.'.  Gdm=*/2mPdm        .        .          (5) 

That  is,  Pm=26r.  Since  at  this  instant  the  load  is  sub- 
jected to  an  upward  force  of  2  G  and  to  a  downward  force 
of  only  G  (gravity)  it  immediately  begins  an  upward  mo- 
tion, reaching  the  point  whence  the  motion  began,  and 
thus  the  oscillation  continues.  We  here  suppose  the  elas- 
ticity of  the  beam  unimpaired.  This  is  called  the  "sud- 
den "  application  of  a  load,  and  produces,  as  shown  above, 
double  the  pressure  on  the  beam  which  it  does  when  grad- 
ually applied,  and  a  double  deflection.  The  work  done 
by  the  beam  in  raising  the  weight  again  is  called  its  re- 
silience. 

Similarly,  if  the  weight  G  is  allowed  to  fall  on  the  mid- 
dle of  the  beam  from  a  height  h,  we  shall  have 


Gx(h+dm),  or  approx.,  Gh,= 
and  hence,  since  (4)  gives  dm  in  terms  of  P 


96  '    -  -  OT  °h  = 


256  MECHANICS   OF  ENGINEERING. 

This  theory  supposes  the  mass  of  the  beam  small  com- 
pared with  the  falling  weight. 

234.  Case  II.    Horizontal  Prismatic  Beam,  Supported  at  Both 
Ends,  Bearing  a  Single  Eccentric  Load.  Weight  of  Beam  Neg- 
lected.— Fig.  227.     The  reactions 
of  the  points  of  support,  P0  and 
PU  are  easily  found  by  consider- 
ing the  whole  beam  free,  and  put- 
ting first  JT(mom.)u  —0,  whence  Pl 
=Pl+ll,a,nd  then  J(mom.)B=0, 
F'«23~-  whence  P0=P(Zl—Z)-5-*i-     P0  and 

P!  will  now  be  treated  as  known  quantities. 

The  elastic  curves  OC  and  CB,  though  having  a  common 
tangent  line  at  C  (and  hence  the  same  slope  «c),  and  a  com- 
mon ordinate  at  C,  have  separate  equations  and  are  both 
referred  to  the  same  origin  and  axes,  as  shown  in  the 
figure.  The  slope  at  0,  «o,  and  that  at  B,alt  are  unknown 
constants,  to  be  determined  in  the  progress  of  the  work. 

Equation  of  OC.  —  Considering  as  free  a  portion  of  the 
beam  extending  from  B  to  a  section  made  anywhere  on 
OC,  x  and  y  being  the  co-ordinates  of  the  neutral  axis  of 
that  section,  we  conceive  the  elastic  forces  put  in  on  the 
exposed  surface,  as  in  the  preceding  problem,  and  put 
2'(mom.  about  neutral  axis  of  the  section)  =0  which  gives 
(remembering  that  here  d-y-^-dx2  is  negative.) 


whence,  by  taking  the  x  anti-derivatives  of  both  members 
El  *L  ^Iv- 


To  find  C,  write  out  this  equation  for  the  point  0,  where 
dy-t-dx=OQ  and  x=0,  and  we  have  C=EIao',  hence  the 
equation  for  slope  is 


FLEXURE  ELASTIC   CURVES.  257 

,      (2) 


Again  taking  the  x  anti-derivatives,  we  have  from  (2) 

Ely  =P        _-P_       +EIa0+(C'=0)     (3) 


(at  Oboth  x  and  y  are  —0  .-.  C"=0).  In  equations  (1),  (2), 
and  (3)  no  value  of  x  is  to  be  used  <0  or  >7,  since  for 
points  in  CB  different  relations  apply,  thus 

Equation  of  CB.  —  Fig.  227.  Let  the  free  body  extend 
from  B  to  a  section  made  anywhere  on  CB.2\moms.),  as 
before,  =0,  gives 


(N.B.  In  (4),  as  in  (1),  ElcPy-r-dx2  is  written  equal  to  a  neg- 
ative quantity  because  itself  essentially  negative  ;  for  the 
curve  is  concave  to  the  axis  X  in  the  first  quadrant  of  the 
co-ordinate  axes.) 

From  (4)  we  have  in  the  ordinary  way  (aj-anti-deriv.) 


(5)-' 


To  determine  C",  consider  that  the  curves  CB  and  OC 
have  the  same  slope  (dy-^dx)  at  C  where  x=l;  hence  put 
a-—  I  in  the  right-hand  members  of  (2)  and  of  (5)'  and 
equate  the  results.  This  gives  C"  =  ^PP-\-EIfj^  and  .'. 


(5) 
(6) 


258  MECHANICS   OF   ENGINEERING. 

At  C,  where  x=l,  both  curves  have  the  same  ordinate  ; 
hence,  by  putting  x=l  in  the  right  members  of  (3)  and  (6)' 
and  equating  results,  we  obtain  C'"=  —  ^PZ3.  .-.  (6)'  be 


(6) 


as  the  Equation  of  CB,  Fig.  227.  But  a^  is  still  an  unknown 
constant,  to  find  which  write  out  (6)  for  the  point  B  where 
x=l,  and  y=0,  whence  we  obtain 


«!=  a  similar  form,  putting  P0  for  P,,  and  (^  —  Z)  for  /. 

235.  Maximum  Deflection  in  Case  IL—  Fig.  227.  The  or- 
dinate ym  of  the  lowest  point  is  thus  found.  Assuming 
£>  /^i»  it  will  occur  in  the  curve  0(7.  Hence  put  the 
dy-^dx  of  that  curve,  as  expressed  in  equation  (2),  =0. 
Also  for  «0  write  its  value  from  (7),  having  putPj=P?-=-Z,, 
and  we  have 


whence    [x  for  max.  y]  = 


Now  substitute  this  value  of  x  in  (3),  also  OQ  from  (7),  and 
^  =Pl-r-li,  whence 


Max.  Deflec.=i/max=V9  .  [P—  3^+2^] 

236.  Case  III.  Horizontal  Prismatic  Beam  Supported  at  Both 
Ends  and  Bearing  a  Uniformly  Distributed  Load  along  its  Whole 
Length.  —  (The  weight  of  the  beam  itself,  if  considered, 


FLEXUKB.  ELASTIC   CUKVES. 


259 


constitutes  a  load  of  this  nature.)  Let  1=  the  length 
of  the  beam  and  w=  the  weight,  per  unit  of  length, 
of  the  loading ;  then  the  load  coming  upon  any  length  x 
will  be  =wx,  and  the  whole  load  =wl.  By  hypothesis  w 
is  constant.  Fig.  228.  From  symmetry  we  know  that  the 


reactions  at  A  and  D  are  each  =j4wl,  that  the  middle  0  of 
the  neutral  line  is  its  lowest  point,  and  the  tangent  line  at 
0  is  horizontal.  Conceiving  a  section  made  at  any  point 
m  of  the  neutral  line  at  a  distance  x  from  0,  consider  as 
free  the  portion  of  beam  on  the  right  of  m.  The  forces 
holding  this  portion  in  equilibrium  are  y£wl,t1o.Q  reaction 
at  B  ',  the  elastic  forces  of  the  exposed  surface  at  m,  viz.: 
the  tensions  and  compressions,  forming  a  couple,  and  J 
the  total  she?r  ;  and  a  portion  of  the  load,  iv(l/J< — x).  The 
sum  of  the  me  tnents  of  these  latter  forces  about  the  neu- 
tral axis  of  m,  is  the  same  as  that  of  their  resultant ;  (i.e., 
their  sum,  since  they  are  parallel),  and  this  resultant  acts  in 
the  middle  of  the  length  y2l — x.  Hence  the  sum  of  these 
moments  =w(y£l — x) ]/2(%l — x).  Now  putting  JT  (mom. 
about  neutral  axis  of  ?ra)=0  for  this  free  body,  we  have 


(i) 


260 


MECHANICS    OF   ENGINEERING. 


Taking  the  a>anti-derivative  of  both  sides  of  (1), 

EITx  =I/2™(I/4l2x-l/^+(c=V  (2) 

as  the  equation  of  slope.  (The  .constant  is  =0  since  at  0 
both  dy-^-dx  and  x  are  =0.)  From  (2), 

^=^(^-Yi2*4)+[C'=0]         .        .     (3) 

A 

which  is  the  equation  of  the  elastic  curve  ;  throughout, 
i.e.,  it  admits  any  value  of  x  from  x=-\-^l  to  x=  —  y2l. 
This  is  an  equation  of  the  fourth  degree,  one  degree  high- 
er than  those  for  the  Curves  of  Cases  I  and  II,  where 
there  were  no  distributed  loads.  If  w  were  not  constant, 
but  proportional  to  the  ordinates  of  an  inclined  right  line, 
eq.  (3)  would  be  of  the  fifth  degree  ;  if  w  were  propor- 
tional to  the  vertical  ordinates  of  a  parabola  with  axis 
vertical,  (3)  would  be  of  the  sixth  degree  ;  and  so  on. 

By  putting  x=l/^l  in  (3)  we  have  the  deflection  of  0  be- 
low the  horizontal  through  A  and  B,  viz.:  (with  W=  total 
load  =wl) 

,      5       «# 
" 


384  "  El     384  '    El 

237.  Case  IV.  Cantilevers.^-A  horizontal  beam  whose  only 
support  consists  in  one  end  being  built  in  a  wall,  as  in 
Fig.  229(a),  or  supported  as  in  Fig. 
229(6)  is  sometimes  called  a  canti- 
lever. Let  the  student  prove  that  in 
Fig.  229(a)  with  a  single  end  load  P, 
the  deflection  of  B  below  the  tangent 
at  Ois  d=^jFVs-r-.£7;the  same  state- 
ment applies  to  Fig.  229(6),  but  the 
tangent  at  0  is  not  horizontal  if  the 
beam  was  originally  so.  It  can  also 
FIG.  229.  "be  proved  that  the  slope  at  B,  Fig. 

229(a)  (from  the  tangent  at  0)  is 


FLEXURE   ELASTIC   CURVES.  261 

PP 


The  greatest  deflection  of  the  elastic  curve  from  the  right 
line  joining  AB,  in  Fig.  229(6),  is  evidently  given  by  the 
equation  for  y  max.  in  §  235,  by  writing,  instead  of  P  of 
that  equation,  the  reaction  at  0  in  Fig.  229(6).  This  assumes 
that  the  max.  deflection  occurs  between  A  and  0.  If  it 
occurs  between  0  and  B  put  (l\—l)  for  I. 

If  in  Fig.  229(a)  the  loading  is  uniformly  distributed 
along  the  beam  at  the  rate  of  w  pounds  per  linear  unit, 
the  student  may  also  prove  that  the  deflection  of  B  below 
the  tangent  at  0  is 


= 

238.  Case  V.  Horizontal  Prismatic  Beam  Bearing  Equal  Ter- 
minal Loads  and  Supported  Symmetrically  at  Two  Points.— 
Fig.  231.  Weight  of  beam  neglected.  In  the  preceding 
cases  we  have  made  use  of  the  approximate  form  ElcPy+dx1 
in  determining  the  forms  of  elastic  curves.  In  the  present 


case  the  elastic  curve  from  0  to  G  is  more  directly  dealt 
with  by  employing  the  more  exact  expression  EI-^p  (see 
§  231)  for  the  moment  of  the  stress-couple  in  any  section. 
The  reactions  at  0  and  C  are  each  =  P,  from  symmetry. 
Considering  free  a  portion  of  the  beam  extending  from  A 
to  any  section  m  between  0  and  C  (Fig.  232)  we  have,  by 
putting  -  (mom.  about  neutral  axis  of  w)=0, 


^62  MECHANICS   OP   ENGINEERING. 

That  is,  the  radius  of  curvature  is  the  same  at  all  points 
of  0(7;  in  other  words  0(7  is  the  arc  of  a  circle  with  the 
above  radius.  The  upward  deflection  of  F  from  the  right 
line  joining  0  and  C  can  easily  be  computed  from  a  knowl- 
edge of  this  fact.  This  is  left  to  the  student  as  also  the 
value  of  the  slope  of  the  tangent  line  at  0  (and  C).  The 
deflection  of  D  from  the  tangent  at  C=l/zP^-7-EI,  as  ID 
Fig.  229(a). 


SAFE  LOADS  IN  FLEXURE. 

239.  Maximum  Moment. — As  we  examine  the  different  sec- 
tions of  a  given  beam  undar  a  given  loading  we  find  differ- 
ent values  of  p,  the  normal  stress  per  unit  of  area  in  the 
outer  element,  as  obtained  from  eq.  (5)  §  229,  viz.: 


.    (1) 


in  which  /  is  the  "  Moment  of  Inertia  "  (§  85)  of  the  plane 
figure  formed  by  the  section,  about  its  neutral  axis,  e  the 
distance  of  the  most  distant  (or  outer)  fibre  from  the  neu- 
tral axis,  and  M  the  sum  of  the  moments,  about  this  neu- 
tral axis,  of  all  the  forces  acting  on  the  free  body  of  which 
the  section  in  question  is  one  end,  exclusive  of  the  stresses 
on  the  exposed  surface  of  that  section.  In  other  words 
M  is  the  sum  of  the  moments  of  the  forces  which  balance 
the  stresses  of  the  section,  these  moments  being  taken 
about  the  neutral  axis  of  the  section  under  examination. 
For  the  prismatic  beams  of  this  chapter  e  and  /  are  the 
same  at  all  sections,  hence  p  varies  with  M  and  becomes  a 
maximum  when  M  is  a  maximum.  In  any  given  case  the 
location  of  the  "  dangerous  section,"  or  section  of  maximum 
M,  and  the  amount  of  that  maximum  value  may  be  deter- 
mined by  inspection  and  trial,  this  being  the  only  method 
(except  by  graphics)  if  the  external  forces  are  detached. 


FLEXURE  SAFE  LOADS.  263 

If,  however,  the  loading  is  continuous  according  to  a  de- 
finite algebraic  law  the  calculus  may  often  be  applied, 
taking  care  to  treat  separately  each  portion  of  the  beam 
between  two  consecutive  reactions  of  supports,  or  detached 
loads. 

As  a  graphical  representation  of  the  values  of  M  along 
the  beam  in  any  given  case,  these  values  may  be  conceived 
laid  off  as  vertical  ordinates  (according  to  some  definite 
scale,  e.g.  so  many  inch-lbs.  of  moment  to  the  linear  inch 
of  paper)  from  a  horizontal  axis  just  below  the  beam.  If 
the  upper  fibres  are  in  compression  in  any  portion  of  the 
beam,  so  that  that  portion  is  convex  downwards,  these  or- 
dinates will  be  laid  off  below  the  axis,  and  vice  versa  ;  for 
it  is  evident  that  at  a  section  where  M=  0,  p  also  =0,  i.e., 
the  character  of  the  normal  stress  in  the  outermost  fibre 
changes  (from  tension  to  compression,  or  vice  versa)  when 
M  changes  sign.  It  is  also  evident  from  eq.  (6)  §  231  that 
the  radius  of  curvature  changes  sign,  and  consequently  the 
curvature  is  reversed,  when  ./If  changes  sign.  These  mo- 
ment ordinates  form  a  Moment  Diagram,  and  the  extremities 
a  Moment  Curve. 

The  maximum  moment,  Mm,  being  found,  in  terms  of 
the  loads  and  reactions,  we  must  make  the  p  of  the  "  dan- 
gerous section,"  where  M=  Mmt  equal  to  a  safe  value  B', 
and  thus  may  write 


(2) 


Eq.  (2)  is  available  for  finding  any  one  unknown  quanti- 
ty, whether  it  be  a  load,  span,  or  some  one  dimension  of 
the  beam,  and  is  concerned  only  with  the  Strength,  and  not 
with  the  stiffness  of  the  beam.  If  it  is  satisfied  in  any 
given  case,  the  normal  stress  on  all  elements  in  all  sections 
is  known  to  be  =  or  <7?',  and  the  design  is  therefore  safe 
in  that  one  respect. 

As  to  danger  arising  from  the  shearing  stresses  in  any 


264 


MECHANICS   OF   ENGINEERING. 


section,  the  consideration  of  the  latter  will  be  taken  up  in 
n  subsequent  chapter  and  will  be  found  to  be  necessary 
only  in  beams  composed  of  a  thin  web  uniting  two  flanges. 
The  total  shear,  however,  denoted  by  J,  bears  to  the  mo- 
ment M,  an  important  relation  of  great  service  in  deter- 
mining Mm.  This  relation,  therefore,  is  presented  in  the 
next  article. 

240,  The  Shear  is  the  First  x-Derivative  of  the  Moment. — 
Fig.  233.  (x  is  the  distance  of  any  section,  measured  parallel 
to  the  beam  from  an  arbitrary 
origin).  Consider  as  free  a  ver- 
tical slice  of  the  beam  included 
between  any  two  consecutive 
vertical  sections  whose  distance 
apart  is  dx.  The  forces  acting 
are  the  elastic  forces  of  the  two 
internal  surfaces  now  laid  bare, 
and,  possibly,  a  portion,  ivdx, 
of  the  loading,  which  at  this 

part  of  the  beam  has  some  intensity  =10  Ibs.  per  running 
linear  unit.  Putting  ^(mom.  about  axis  .AT')=0  we  have 
(noting  that  since  the  tensions  and  compressions  of  section 
N  form  a  couple,  the  sum  of  their  moments  about  N'  is 
just  the  same  as  about  Nt) 


pdF 


But  P-  =  M,  the  Moment  of  the  left  hand  section,^  =  M>  , 


that  of  the  right  ;   whence  we   may  write,  after  dividing 
through  by  dx  and  transposing, 


2 


dM 

-,— 

dx 


T 

=±.7; 


(3) 


for  10  If  vanishes  when  added  to  the  finite  J,  and  M' — M= 
dM=  increment  of  the  moment  corresponding  to  the  incre- 
ment, dx,  of  x.  This  proves  the  theorem. 


FLEXURE.  SAFE  LOADS.  205 

Now  the  value  of  x  which  renders  M  a  maximum  or 
minimum  would  be  obtained  by  putting  the  derivative 
dM--rdx=  zero ;  hence  we  may  state  as  a 

Corollary. —At  sections  where  the  moment  is  a  maximum 
or  minimum  the  shear  is  zero. 

The  shear  J  at  any  section  is  easily  determined  by  con- 
sidering free  the  portion  of  beam  from  the  section  to  either 
end  of  the  beam  and  putting  ^'(vertical  components) =0. 

In  this  article  the  words  maximum  and  minimum  are 
used  in  the  same  sense  as  in  calculus ;  i.e.,  graphically, 
they  are  the  ordinates  of  the  moment  curve  at  points 
where  the  tangent  line  is  horizontal.  If  the  moment  curve  be 
reduced  to  a  straight  line,  or  a  series  of  straight  lines,  it 
has  no  maximum  or  minimum  in  the  strict  sense  just 
stated  ;  nevertheless  the  relation  is  still  practically  borne 
out  by  the  fact  that  at  the  sections  of  greatest  and  least 
ordinates  in  the  moment  diagram  the  shear  changes  sign 
suddenly.  This  is  best  shown  by  drawing  a  shear  diagram, 
whose  ordinates  are  laid  off  vertically  from  a  horizontal 
axis  and  under  the  respective  sections  of  the  beam.  They 
will  be  laid  off  upward  or  downward  according  as  «7is 
found  to  be  upward  or  downward,  when  the  free  body  con- 
sidered extends  from  the  section  toward  the  right. 

In  these  diagrams  the  moment  ordinates  are  set  off  on 
an  arbitrary  scale  of  so  many  inch-pounds,  or  foot-pounds, 
to  the  linear  inch  of  paper ;  the  shears  being  simply 
pounds,  or  some  other  unit  of  force,  on  a  scale  of  so  many 
pounds  to  the  inch  of  paper.  The  scale  on  which  the 
beam  is  drawn  is  so  many  feet,  or  inches,  to  the  inch  of 
paper. 

241.  Safe  Load  at  the  Middle  of  a  Prismatic  Beam  Support- 
ed at  the  Ends. — Fig.  234.  The  reaction  at  each  support 
is  *4P.  Make  a  section  n  at  any  distance  x<±  from  B. 
Consider  the  portion  nB  free,  putting  in  the  proper  elas- 
tic and  external  forces.  The  weight  of  beam  is  neglected. 
T?rorn  ^(mom.  about  w)=0  we  have 


26G 


MECHANICS   OF   ENGINEERING. 


P    =    x;Le.t  M=%Px 

Q          A 

Evidently  M  is  proportional  to  x,  and  the  ordinates  repre- 
senting it  will  therefore  be  limited  by  the  straight  line 


B'R,  forming  a  triangle  B'EA.  From  symmetry,  another 
triangle  0  RA'  forms  the  other  half  of  the  moment  dia- 
gram. From  inspection,  the  maximum  M  is  seen  to  be  in 
the  middle  where  x=  y2l,  and  hence 


•    (1) 


Again  by  putting  2Yvert.  compons.)=0,  for  the  free  body 
nB  we  have 


J= 


and  must  point  downward  since  ~  points  upward.  Hence 
the  shear  is  constant  and  =  y2P  at  any  section  in  the  right 
hand  half.  If  n  be  taken  in  the  left  half  we  would  have, 
nB  being  free,  from  ^'(vert.  com.)=0, 


FLEXURE.  SAFE  LOADS.  267 

the  same  numerical  value  as  before  ;  but  e/must  point  up- 
ward, since  ~  at  B  and  J  at  n  must  balance  the  downward 
P  at  A.  At  A,  then,  the  shear  changes  sign  suddenly, 
that  is,  passes  through  the  value  zero;  also  a,i  A,  M  is  a 
maximum,  thus  illustrating  the  statement  in  §  240.  Notice 
the  shear  diagram  in  Fig.  234. 

To  find  the  safe  load  in  this  case  we  write  the  maximum 
value  of  the  normal  stress,  p,=  R',  a  safe  value,  (see  table 
in  a  subsequent  article)  and  solve  the  equation  for  P. 
But  the  maximum  value  of  p  is  in  the  outer  fibre  at  A, 
since  Jf  for  that  section  is  a  maximum.  Hence 


(2) 


is  the  equation  for  safe  loading  in  this  case,  so  far  as  the 
normal  stresses  in  any  section  are  concerned. 

EXAMPLE.  —  If  the  beam  is  of  wood  and  has  a  rectangu- 
lar section  with  width  b=  2  in.,  height  h-*=  4  in.,  while  its 
length  Z=  10  ft.,  required  the  safe  load,  if  the  greatest  nor- 
mal stress  is  limited  to  1,000  Ibs.  per  sq.  in.  Use  the 
pound  and  inch.  From  §  90  I=>l/a  ^3=Yi2X2x64=10.66 
biquad.  inches,  while  e=l=2  in. 


.  p== 


4x1,000x10.66  _ 


le  120x2 


242.  Safe  Load  Uniformly  Distributed  along  a  Prismatic  Beam 
Supported  at  the  Ends. —Let  the  load  per  lineal  unit  of  the 
length  of  beam  be  =w>  (this  can  be  made  to  include  the 
weight  of  the  beam  itself).  Fig.  235.  From  symmetry, 

each  reaction  =  }4wl.     For  the  free  body  nO  we  have,  put- 
ting <T(mom.  about  w)=0,  ' 


268 


MECHANICS    OF   ENGINEERING. 


which  gives  M  for  any  section  by  making  x  vary  from  0 
to  Z.  Notice  that  in  this  case  the  law  of  loading  is  con- 
tinuous along  the  whole  length,  and  that  hence  the  mo- 
ment curve  is  continuous  for  the  whole  length. 


w-  wl 


To  find  the  shear  J,  at  n,  we  may  either  put  ^(vert.  com- 
pons.)=0  for  the  free  body,  whence  J=  ^wl — wx,  and  must 
therefore  be  downward  for  a  small  value  of  x  ;  or,  employ- 
ing §  240,  we  may  write  out  dM-^dx,  which  gives 


dM 


(i) 


the  same  as  before.  To  find  the  max.  M,  or  Mm,  put  J=Q, 
which  gives  x=^y2l.  This  indicates  a  maximum,  for  when 
substituted  in  d^M-^dx2,  i.e.,  in  — w,  a  negative  result  is 
obtained.  Hence  Mm  occurs  at  the  middle  of  the  beam  and 
its  value  is 


=«#=    m 


(2) 


the  equation  of  safe  loading.      W=  total  load=«?Z. 

It  can  easily  be  shown  that  the  moment  curve  is  a  por- 


FLEXURE.  SAFE  LOADS.  269 

jion  of  a,  parabola,  whose  vertex  is  at  A"  under  the  mid- 
dle of  the  beam,  and  axis  vertical.  The  shear  diagram 
consists  of  ordinates  to  a  single  straight  line  inclined  to 
its  axis  and  crossing  it,  i.e.,  giving  a  zero  shear,  under  the 
middle  of  the  beam,  where  we  find  the  max.  M. 

If  a  frictionless  dove-tail  joint  with  vertical  faces  were 
introduced  at  any  locality  in  the  beam  and  thus  divided 
the  beam  into  two  parts,  the  presence  of  J  would  be  made 
manifest  by  the  downward  slipping  of  the  left  hand  part 
on  the  right  hand  part  if  the  joint  were  on  the  right  of  the 
middle,  and  vice  versa  if  it  were  on  the  left  of  the  middle. 
This  shows  why  the  ordinates  in  the  two  halves  of  the 
shear  diagram  have  opposite  signs.  The  greatest  shear 
is  close  to  either  support  and  is  Jm= 


243.  Prismatic  Beam  Supported  at  its  Extremities  and  Loaded 
in  any  Manner.    Equation  for  Safe  Loading. — Fig.  236.    Given 
p          p2          Pj  the  loads  P,,  P2,  and  P3,  whose 

B  0    distances  from  the  right  sup- 

*       iLj  port  are  11}  12,  and  4 ;  ,required 
-!-p  the  equation  for  safe  loading  ; 
i.e.,    find   Mm    and   write   it    = 
Rl+e. 

If  the  moment  curve  were 
continuous,  i.e.,  if  M  were  a 
continuous  function  of  x  from 
end  to  end  of  the  beam,  we 
could  easily  find  Mm  by  making 
dM-r-dx=Q,  i.e.,  J=0,  and  sub- 
stitute the  resulting  value  of  x  in  the  expression  for  M. 
But  in  the  present  case  of  detached  loads,  J  is  not  zero, 
necessarily,  at  any  section  of  the  beam.  Still  there  is 
some  one  section  where  it  changes  sign,  i.e.,  passes  sud- 
denly through  the  value  zero,  and  this  will  be  the  section 
of  greatest  moment  (though  not  a  maximum  in  the  strict 
sense  used  in  calculus).  By  considering  any  portion  n^ 
as  free,  «7is  found  equal  to  the  Reaction  at  0  Diminished  by 
the  Loads  Occuring  Between  n  and  0.  The  reaction  at  B  is 


MECHANICS   OF   ENGINEERING. 


obtained  by  treating  the  whole  beam  as  free  (in  which  case 
no  elastic  forces    come  into   play)    and  putting  ,T(moni. 
about  O)=0;  while  that  at  0,=P0=Pl+P2-\-Ps—  PB 
If  n  is  taken  anywhere  between  0  and  E,  J=P0 

E    "    F,J=P0-P, 
F    "    H,  J^Py-P^Pi 
H    "    B,  J=P0-PV-P2-P3 

This  last  value  of.  J  also  =  the  reaction  at  the  other 
support  ,  B.  Accordingly,  the  shear  diagram  is  seen  to 
consist  of  a  number  of  horizontal  steps.  The  relation 
J=dM-7-dx  is  such  that  the  slope  of  the  moment  curve  is 
proportional  to  the  ordinate  of  the  shear  diagram,  and 
that  for  a  sudden  change  in  the  slope  of  the  moment  curve 
there  is  a  sudden  change  in  the  shear  ordinate.  Hence  in 
the  present  instance,  J  being  constant  between  any  two 
consecutive  loads,  the  moment  curve  reduces  to  a  straight 
line  between  the  same  loads,  this  line  having  a  different 
inclination  under  each  of  the  portions  into  which  the  beam 
is  divided  by  the  loads.  Under  each  load  the  slope  of  the 
moment  curve  and  the  ordinate  of  the  shear  diagram  change 
suddenly.  In  Fig.  236  the  shear  passes  through  the  value 
zero,  i.e.,  changes  sign,  at  F;  or  algebraically  we  are  sup- 
posed to  find  that  P0-Pl  is  +  while  P0-Pl-P2  is  -,  in 
the  present  case.  Considering  FO,  then,  as  free,  we  find 
Mm  to  be 

Mm=P0lt—Pi(li—li)  and  the  equation  for  safe  loading  is 

i)  (1) 


(i.e.,  if  the  max.  M  is  at  F).  It  is  also  evident  that  the 
greatest  shear  is  equal  to  the  reaction  at  one  or  the  other 
support,  whichever  is  the  greater,  and  that  the  moment 
at  either  support  is  zero. 

The  student  should  not  confuse  the  moment  curve,  which 


FLEXUKE.  SAFE   LOADS. 


273 


is  entirely  imaginary,  with  the  neutral  line  (or  elastic 
curve)  of  the  beam  itself.  The  greatest  moment  is  not 
necessarily  at  the  section  of  maximum  deflection  of  the 
neutral  line  (or  eJastic  curve). 

For  the  case  in  Fig.  236  we  may  therefore  state  that  the 
max.  moment,  and  consequently  the  greatest  tension  or 
compression  in  the  outer  fibre,  will  be  found  in  the  sec- 
tion under  that  load  for  which  the  sum  of  the  loads  (in- 
cluding this  load  itself)  between  it  and  either  support  first 
equals  or  exceeds  the  reaction  of  that  support.  The 
amount  of  this  moment  is  then  obtained  by  treating  as  free 
either  of  the  two  portions  of  the  beam  into  which  this 
section  divides  the  beam. 

244.  Numerical  Example  of  the  Preceding  Article. — Fig.  237. 
Given  Plt  P2,  P3,  equal  to  }4  ton,  1  ton,  and  4  tons,  re- 


spectively ;  ^  =5  feet,  £2=  7  feet,  and  ^=  10  feet ;  while  the 
total  length  is  15  feet.  The  beam  is  of  timber,  of  rectan- 
gular cross-section,  the  horizontal  width  being  6=10 
inches,  and  the  value  of  R'  (greatest  safe  normal  stress), 
=  ^  ton  per  sq.  inch,  or  1,000  Ibs.  per  sq  inch. 


272  MECHANICS   OF   ENGINEERING. 

Required  the  proper  depth  h  lor  the  beam,  for  safe  load- 
ing- 
Solution. — Adopting  a  definite  system  of  units,  viz.,  the 
inch-ton-second  system,  we  must  reduce  all  distances  such 
as  I,  etc.,  to  inches,  express  all  forces  in  tons,  write  R'  =  *4 
(tons  per  sq.  inch),  and  interpret  all  results  by  the  same  sys- 
tem. Moments  will  be  in  inch-tons,  and  shears  in  tons. 
[N.  B.  In  problems  involving  the  strength  of  materials 
the  inch  is  more  convenient  as  a  linear  unit  than  the  foot, 
since  any  stress  expressed  in  Ibs.,  or  tons,  per  sq.  inch,  is 
numerically  144  times  as  small  as  if  referred  to  the  square 
foot.] 

Making  the  whole  beam  free,  we  have  from  moms,  about 
O,  Pfl=^  [Xx60+lx84+4xl20]=3.3  tons.-.  P0=5.5— 
3.3=2.2  tons. 

The  shear  anywhere  between  0  and  jE7is  J=  P0=2.2  tons. 

E  and  F  is  J  =2.2— X  =1.7 

tons. 
The  shear  anywhere  between  F  and  NIB  J  =2.2 — ^ — 1  = 

0.7  tons. 
The  shear  anywhere  between  //  and  £  is  J  =  2.2 — */£ — 1 

— 4= — 3.3  tons. 

Since  the  shear  changes  sign  on  passing  H,  .:  the  max. 
moment  is  at  H',  whence  making  HO  free,  we  have 
M  at  H=Mm  =2.2  x  120—^  X  60—1  x  36 =198  inch-tons. 

T>l  T 

For  safety  Mra  must  =  — ,  in   which   R  =  ^    ton   per   sq. 
c 

inch,  e=i^h  =  ^  of  unknown  depth  of  beam,  and  /,  §90,  = 

I  b h3,  with  6=10  inches 

,:^.  y2  .jLxlOA3=198;  or  A2=237.6/.  A=15.4  inches. 

245.  Comparative  Strength  of  Rectangular  Beams. — For  such 
a  beam,  under  a  given  loading,  the  equation  for  safe  load- 

^  ^  «T 

—  =  .«;„  i.  e.  %  R  bW=Mn (1) 


FLEXURE.   SAFE  LOADS.  273 

whence  the  following  is  evident,  (since  for  the  same  length, 
mode  of  support,  and  distribution  of  load,  Mm  is  propor- 
tional to  the  safe  loading.) 

For  rectangular  prismatic  beams  of  the  same  length, 
same  material,  same  mode  of  support  and  same  arrange- 
ment of  load  : 

(1)  The  safe  load  is  proportional  to  the  width  of  beams 
'having  the  same  depth  (A). 

(2)  The  safe  load  is  proportional  to  the  square  of  the 
depth  of  beams  having  the  same  width  (6). 

(3)  The  safe  load  is  proportional  to  the  depth  of  beams 
having  the  same  volume  (i.  e.  the  same  bh). 

(It  is  understood  that  the  sides  of  the  section  are  hori- 
zontal and  vertical  respectively  and  that  the  material  is 
homogeneous.) 

246.  Comparative  Stiffness  of  Rectangular  Beams.— Taking  the 
deflection  under  the  same  loading  as  an  inverse  measure 
of  the  stiffness,  and  noting  that  in  §§  233,  235,  and  236, 
this    deflection   is    inversely  'proportional    to    I=^bhs  = 
the  "  moment  of  inertia  "  of  the  section  about  its  neutral 
axis,  we  may  state  that : 

For  rectangular  prismatic  beams  of  the  same  length, 
same  material,  same  mode  of  support,  and  same  loading  : 

(1)  The  stiffness  is  proportional  to  the  width  for  beams 
of  the  same  depth. 

(2)  The  stiffness   is   proportional   to  the  cube  of  the 
height  for  beams  of  the  same  width  (6). 

(3)  The  stiffness  is  proportional  to  the  square  of  the 
depth  for  beams  of  equal  volume  (bJil). 

(4)  If  the  length  alone  vary,  the  stiffness  is  inversely 
proportional  to  the  cube  of  the  length. 

247.  Table  of  Moments  of  Inertia. — These  are  here  recapitu- 
lated for  the  simpler  cases,  and  also  the  values  of  e.  the 
distance  of  the  outermost  fibre  from  the  axis. 

Since  the  stiffness  varies  as  /(other  things  being  equal), 


274 


MECHANICS   OP   ENGINEERING. 


while  the  strength  varies  as  I-t-e,  it  is  evident  that  a 
square  beam  has  the  same  stiffness  in  any  position  (§89), 
while  its  strength  is  greatest  with  one  side  horizontal,  for 
then  e  is  smallest,  being  =y^b. 

Since  for  any  cross-section  7=  CdF  z\  in  which  2=the 

distance  of  any  element,  dF,  of  area  from  the  neutral  axis, 
a  beam  is  made  both  stiffer  and  stronger  by  throwing 
most  of  its  material  into  two  flanges  united  by  a  vertical 
web,  thus  forming  a  so-called  "  I-beam  "  of  an  I  shape.  But 
not  without  limit,  for  the  web  must  be  thick  enough  to 
cause  the  flanges  to  act  together  as  a  solid  of  continuous 
substance,  and,  if  too  high,  is  liable  to  buckle  sideways, 
thus  requiring  lateral  stiffening.  These  points  will  ba 
treated  later. 


SECTION. 

I 

6 

Bectangle,  width  =  b,  depth  =  h  (vertical) 

Vn  M« 

%h 

Hollow  Rectangle,  symmet.  about  neutral  axis.     See  1 
Fig.  238  (a)                                                                       ) 

Via  [»i  Ai*-68  A»J 

%ht 

Triangle,  width  =&,  height  =  h,  neutral  axis  parallel  1 
to  base  (horizontal).                                                            1 

Vie  M« 

%h 

Circle  of  radius  r 

K*r« 

r 

Ring  of  concentric  circles.    Pig.  238  (b) 

X*(r*  ,—  r«t) 

?i 

Rhombus;  Fig.  238  (c)  h  =  diagonal  which  is  vertical. 

»/48  6A» 

Xh. 

Square  with  side  b  vertical. 

Vu  !>* 

y*b 

"      "     6at45°withhoriz. 

Vis  *4 

WV2 

248.    Moment   of  Inertia  of  I-beams,   Box-girders,    Etc. — In 
common  with  other  large  companies,  the  N.  J.  Steel  and 


FLEXURE.      SATE   LOADS. 


275 


Iron  Co.  of  Trenton,  N.  J.  (Cooper,  Hewitt  &  Co.)  manu- 
facture prismatic  rolled  beams  of  wrought-iron  variously 
called  /-beams,  deck -beams,  rails,  and  "  shape  iron,"  (in- 
cluding channels,  angles,  tees,  etc.,  according  to  the  form 
of  section.)  See  fig.  239  for  these  forms.  The  company 


ilL 


T 


CHANNEL.     DECK-BEAM-. 

FIG.  239. 


publishes  a  pocket-book  giving  tables  of  quantities  rela- 
ting to  the  strength  and  stiffness  of  beams,  such  as  the 
safe  loads  for  various  spans,  moments  of  inertia  of  their 
sections  in  various  positions,  etc.,  etc.  The  moments  of 
inertia  of  /-beams  and  deck-beams  are  computed  accord- 
ing to  §§  92  and  93,  with  the  inch  as  linear  unit.  The 
/-beams  range  from  4  in.  to  20  inches  deep,  the  deck- 
beams  being  about  7  and  8  in.  deep. 

For  beams  of  still  greater  stiffness  and  strength  com- 
binations of  plates,  channels,  angles,  etc.,  are  riveted  to- 
gether, forming  "  built-beams,::  or  "  plate  girders."  The 
proper  design  for  the  riveting  of  such  beams  will  be  ex- 
amined later.  For  the  present  the  parts  are  assumed  to 
act  together  as  a  continuous  mass.  For  example,  Fig.  240 
shows  a  "  box-girder,"  formed  of  two  "  channels "  and 
two  plates  riveted  together.  If  the  axis  of  symmetry,  N, 
is  to  be  horizontal  it  becomes  the  neu- 
tral  ax^s-  -ket  @=  the  moment  of  iner- 
tia of  one  channel  (as  given  in  the 
pocket-book  mentioned)  about  the  axis 
L_^  N  perpendicular  to  the  web  of  the  chan- 
HT  '  \P  nel-  Then  the  total  moment  of  inertia  oj 

FIG.  340.  the  combination  is  (nearly) 


(1) 


27G 


MECHANICS   OF    ENGINEERING. 


In  (1),  b,  t,  and  d  are  the  distances  given  in  Fig.  240  (d  ex- 
tends to  the  middle  of  plate)  while  d'  and  t'  are  the  length 
and  width  of  a  rivet,  the  former  from  head  to  head 
(i.e.,  d'  and  t'  are  the  dimensions  of  a  rivet-hole). 

For  example,  a  box-girder  of  wrought-iron  is  formed  of 
two  15-inch  channels  and  two  plates  10  inches  wide  and  1 
inch  thick,  the  rivet  holes  ^  in.  wide  and  1^  in.  long. 
That  is,  6=10;  t=l;  d=8;  t'  =  %-,  and  d'=l%  inches. 
Also  from  the  pocket-book  we  find  that  for  the  channel  in 
question,  (7=376  biquadratic  inches.  Hence,  eq.  (1) 

IN  =  752+2xlOxlx64— 4x^X^(8— X)2=1737biquadr.in. 

Also,  since  in  this  instance  e  =  8^  inches,  and  12000 
Ibs.  per  sq.  inch  (or  6  tons  per  sq.  in.)  is  the  value  for  R' 
(=greatest  safe  normal  stress  en  the  outer  element  of  any 
cross-section)  used  by  the  Trenton  Co.  (for  wrought  iron), 

we  have  ^1^X1737=^,00  inch-lbs.  ' 

That  is,  the  box -girder  can  safely  bear  a  maximum  mo- 
ment, Mm,  =  2451700  inch-lbs.  =  1225.8  inch-tons,  as  far 
as  the  normal  stresses  in  any  section  are  concerned. 
(Proper  provision  for  the  shearing  stresses  in  the  section, 
and  in  the  rivets,  will  be  considered  later). 

249.  Strength  of  Cantilevers. — In   Fig.   241  with    a  single 
w-w?  concentrated  load  P  at  the 

projecting    extremity,   we 
J°  easily  find  the  moment  at 
,  to  be  M  =Px,  and  the 
/  max.  moment  to    occur  at 


the  section  next  the  wall, 
{f  its  value  being  Mm=PL 

The  shear,  J,  is  constant, 
and  =  P  at  all  sections. 
The  moment  and  shear  diagrams  are  drawn  in  accordance 
with  these  results. 


FLEXURE.      SAFE   LOADS. 


If  the  load  W  =  id  is  uniformly  distributed  on  the  can- 
tilever, as  in  fig.  242,  by  making  nO  free  we  have,  putting 
.  about  n)  =  0, 


=tfwP=  */2  Wl. 


P--=wx  .  ~  .-. 

Hence  the  moment  curve  is  a  parabola,  whose  vertex  is  at 
O7  and  axis  vertical.  Putting  I  (vert,  compons.)  =  0  we 
obtain  J  =  wx.  Hence  the  shear  diagram  is  a  triangle, 
and  the  max.  J=  wl  =  W. 

250.  Resume"  of  the  Four  Simple  Cases.  —  The  following  table 
shows  the  values  of  the  deflections  under  an  arbitrary 
load  P,  or  W,  (within  elastic  limit),  and  of  the  safe  load  ; 


Cantilevers  . 

Beams  with  two  end  supports. 

With  one  end 
loadP 

Fi-   ail 

With  unif.  load 
W=  wl 
Fig.  242 

Load  Pin 
middle 
Fig.  234 

Unif.  load 
W=wl 
Fig.  235 

Deflection 

H.g 

/8'lli 

1    PI* 

5      WP 
384'  El 

(  Safe  load  (from  ?J[ 

1  =  Xm) 

ifi 

le 

R'l 

*ar 

8^T 

Relative  strength 

I 

2 

4 

8 

j  Relative  stiffness 
'(  under  same  load 

(  Relative  stiffness 
'(  under  safe  load 

1 
1 

•A 

16 
4 

16 
5 

(Max.  shear  =  Jm,  (and 
1  location, 

P,  (at  wall) 

W,  (atwaU) 

HP,  (at  supp). 

X  W,  (at  supp). 

also  the  relative  strength,  the  relative  stiffness  (under  the 
same  load),  and  the  relative  stiffness  under  the  safe  load, 
for  the  same  beam. 

The  max.  shear  will  be  used  to  determine  the  proper 
web-thickness  for  /-beams  and  "  built-girders."  The  stu- 
dent should  carefully  study  the  foregoing  table,  noting 
especially  the  relative  strength,  stiffness,  and  stiffness 
under  safe  load,  of  the  same  beam. 

Thus,  a  beam  with  two  end  supports  will  bear  a  double 


278  MECHANICS   OF   ENGINEERING. 

load,  if  uniformly  distributed  instead  of  concentrated  in 
the  middle,  but  will  deflect  ^  more ;  whereas  with  a  given 
load  uniformly  distributed  the  deflection  would  be  only 
^  of  that  caused  by  the  same  load  in  the  middle,  provided 
the  elastic  limit  is  not  surpassed  ii?  either  case. 

251.  R',  etc.  For  Various  Materials.— The  iormula^2/=  Mmt 

e 

from  which  in  any  given  case  of  flexure  we  can  compute 
the  value  of  pm,  the  greatest  normal  stress  in  any  outer 
element,  provided  all  the  other  quantities  are  known, 
holds  good  theoretically  within  the  elastic  limit  only. 
Still,  some  experimenters  have  used  this  formula  for  the 
rupture  of  beams  by  flexure,  calling  the  value  of  pm  thus 
obtained  the  Modulus  of  Rupture,  R.  R  may  be  found  to 
differ  considerably  from  both  the  T  or  C  of  §  203  with 
some  materials  and  forms,  being  frequently  much  larger. 
This  might  be  expected,  since  even  supposing  the  relative 
extension  or  compression  (i.e.,  strain)  of  the  fibres  to  be 
proportional  to  their  distances  from  the  neutral  axis  as 
the  load  increases  toward  rupture,  the  corresponding 
stresses,  not  being  proportional  to  these  strains  beyond  the 
elastic  limit,  no  longer  vary  directly  as  the  distances  from  the 
neutral  axis ;  and  the  neutral  axis  does  not  pass  through  the 
centre  of  gravity  of  the  section,  necessarily. 

The  following  table  gives  average  values  for  R,  R,  R", 
and  E  for  the  ordinary  materials  of  construction.  E,  the 
modulus  of  elasticity  for  use  in  the  formulae  for  deflection, 
is  given  as  computed  from  experiments  in  flexure,  and  is 
nearly  the  same  as  Et  and  Ec. 

In  any  example  involving  R',  e  is  usually  written  equal 
to  the  distance  of  the  outer  fibre  from  the  neutral  axis, 
whether  that  fibre  is  to  be  in  tension  or  compression ; 
since  in  most  materials  not  only  is  the  tensile  equal  to  the 
compressive  stress  for  a  given  strain  (relative  extension 
or  contraction)  but  the  elastic  limit  is  reached  at  about 
the  same  strain  both  in  tension  and  compression. 


FLEXURE.       SAFE  LOADS.  279 

TABLE  FOB  USE  IN  EXAMPLES  IN  FLEXURE. 


Timber. 

Cast  Iron. 

Wro't  Iron. 

Steel. 

Max.  safe  stress  in  outer  fi-  ) 
V        1,000 
bre  -.R'dbs.  per  sq.  inch).  J 

6,000  in  tens. 
12,000  in  comp. 

12,000 

15,000 
to 
40,000 

Stress  in  outer  fibre  at  Has.  | 
limit  -J^/Ubs.  per  sq.  In.)  f 

17.000* 
to 
35,000 

30,000 
and  upward. 

"  Modul.  of  Rupture  " 
=-.ff-=lbs.  per  sq.  inch. 
#=Mod.  of  Elasticity, 
=lbs.  per  sq.  inch. 

)          4,000 
V            to 
\      «0,000 

)  1,000,000 
V       to 
j  3,000,000 

40,000 
17,000,000 

50,000 
•  25,000,000 

120,000 
Hard  Steel. 
30,000,000 

In  the  case  of  cast  iron,  however,  (see  §  203)  the  elastic 
limit  is  reached  in  tension  with  a  stress  =9,000  Ibs.  per 
sq.  inch  and  a  relative  extension  of  ^  of  one  per  cent., 
while  in  compression  the  stress  must  be  about  double  to 
reach  the  elastic  limit,  the  relative  change  of  form  (strain) 
being  also  double.  Hence  with  cast  iron  beams,  once 
largely  used  but  now  almost  entirely  displaced  by  rolled 
wrought  iron  beams,  an  economy  of  material  was  effected 
by  making  the  outer  fibre  on  the  compressed  side  twice 
as  far  from  the  neutral  axis  as  that  on  the  stretched  side. 
Thus,  Fig.  243,  cross-sections  with  unequal  flanges  were 
used,  so  proportioned  that  the  centre  of 
gravity  was  twice  as  near  to  the  outer 
fibre  in  tension  as  to  that  in  compression, 
i.e.,  e?=2ei ;  in  other  words  more  material 
is  placed  in  tension  than  in  compression. 
FIG.  m  The  fibre  A  being  in  tengion  (within  elas- 

tic limit),  that  at  B,  since  it  is  twice  as  far  from  the  neu- 
tral axis  and  on  the  other  side,  is  contracted  twice  as  much 
as  A  is  extended  ;  i.e.,  is  under  a  compressive  strain 
double  the  tensile  strain  at  A,  but  in  accordance  with  the 
above  figures  its  state  of  stress  is  proportionally  as  much 
within  the  elastic  limit  as  that  of  A. 

Steel  beams  are  gradually  coming  into  use,  and  may  ul- 
timately replace  those  of  wrought  iron. 

*  In  the  tests  by  U.  S.  Gov.  in  1879  with  I-beams.  R"  ranged  from  25,000 
to  38,000,  and  tlie  elastic  limit  was  reached  with  less  stress  in  the  Innre 
than  in  the  smaller  beams.  Also,  for  the  same  beam,  R'  decreased  with 
larger  spans. 


280  MECHANICS   OF   ENGINEERING. 

The  great  range  of  values  of  E  for  timber  is  due  noi 
only  to  the  fact  that  the  various  kinds  of  wood  differ 
widely  in  strength,  while  the  behavior  of  specimens  of 
any  one  kind  depends  somewhat  on  age,  seasoning,  etc., 
but  also  to  the  circumstance  that  the  size  of  the  beam  un- 
der experiment  Las  much  to  do  with  the  result.  The  ex- 
periments of  Prof.  Lanza  at  the  Mass.  Institute  of  Tech- 
nology in  1881  were  made  on  full  size  lumber  (spruce),  of 
dimensions  such  as  are  usually  tak^n  for  floor  beams  in 
buildings,  and  gave  much  smaller  values  of  R  (from  3,200 
to  8,700  Ibs.  per  sq.  inch)  than  had  previously  been  ob- 
tained. The  loading  employed  was  in  most  cases  a  con- 
centrated load  midway  between  the  two  supports. 

These  low  values  are  probably  due  to  the  fact  that  in 
large  specimens  of  ordinary  lumber  the  continuity  of  it& 
substance  is  more  or  less  broken  by  cracks,  knots,  etc., 
the  higher  values  of  most  other  experimenters  having 
been  obtained  with  small,  straight-grained,  selected  pieces, 
from  one  foot  to  six  feet  in  length. 

The  value  R'  =12,000  rbs.  per  sq.  inch*  is  employed  by 
the  N.  J.  Iron  and  Steel  Co.  in  computing  the  safe  loads 
for  their  rolled  wrought  iron  beams,  with  the  stipulation 
that  the  beams  (which  are  high  and  of  narrow  width)  must 
be  secure  against  yielding  sideways.  If  such  is  not  the 
case  the  ratio  of  the  actual  safe  load  to  that  computed  with 
72' =12,000  is  taken  less  and  less  as  the  span  increases. 
The  lateral  security  referred  to  may  be  furnished  by  the 
brick  arch-filling  of  a  fire-proof  floor,  or  by  light  lateral 
bracing  with  the  other  beams. 

252.  Numerical  Examples. — EXAMPLE  1. — A  square  bar  of 
wrought  iron,  1*^  in.  in  thickness  is  bent  into  a  circular 
arc  whose  radius  is  200  ft.,  the  plane  of  bending  being  par- 
allel to  the  side  of  the  square.  Required  the  greatest  nor- 
mal stress  pm  in  any  outer  fibre. 

Solution.     From  §§  230  and  231  we  may  write 

— t  =P—  .•.  p=eE-r-n,  i.e.,  is  constant. 
P         e 

*  For  '.heir  soft-steel  bc;ims  this  com|>:my  uses  16,000  Ibs.  persq.  inch. 


FLEXURE.      SAFE   LOADS.  '281 

For  the  units  inch  and  pound  (viz.  those  of  the  table  in  § 
251)  we  have  e=%  in.,  p  =2,400  in.,  and  ^=25,000,000  Ibs. 
per  sq.  inch,  anl  .-. 

p=pm  =  ^X 25,000,000-^ 2,400 =7,812  Ibs.  per  sq.  in., 

which  is  quite  safe.  At  a  distance  of  }4  inch  from  the 
neutral  axis,  the  normal  stress  is  =^-^-^pm  =  2/^pm= 
5,208  Ibs.  per  sq.  in.  (If  the  force-plane  (i.e.,  plane  of 
bending)  were  parallel  to  the  diagonal  of  the  square,  e 
would  =l/2X  1.5^2  inches,  giving  jt>m  =  [  7,812x^2  ]  Ibs. 
per  sq.  in.)  §  238  shows  an  instance  where  a  portion,  OC, 
Fig.  231,  is  bent  in  a  circular  arc. 

EXAMPLE  2. — A  hollow  cylindrical  cast-iron  pipe  of  radii 
3 '/2  and  4  inches*  is  supported  at  its  ends  and  loaded  in 
middle  (see  Fig.  234).  Required  the  safe  load,  neglecting 
the  weight  of  the  pipe.  From  the  table  in  §  250  we  have 
for  safety 

.        ***£ 

From  §  251  we  put  #=6,000  Ibs.  per  sq.  in.;  and  from  § 
247/=*-(r,4 — r24);  and  with  these  values,  r2  being  =4»  r\  — 
4}  e— n=4,  ~=-y-  and  Z=144  inches  (the  inch  must  "be  the 
unit  of  length  since  .#'  =  6,000  Ibs.  per  sq.  inch)  we  have 

P=4x6,000x^-  -5- (256-  150)-r- [144x4]  .-.  P= 3,470  Ibs. 
The  weight  of  the  beam  itself  is  G=  7?,  (§  7),  i.e., 

G=-(r?-r*}lr=  f (16-12^)144x^=443  Ibs. 

(Notice  that  f,  here,  must  be  Ibs.,  per  cubic  inch}.  This 
weight  being  a  uniformly  distributed  load  is  equivalent  to 
half  as  much,  221  Ibs.,  applied  in  the  middle,  as  far  as  the 
strength  of  the  beam  is  concerned  (see  §  250),  .'.  P  must  be 
taken  =3,249  Ibs.  when  the  weight  of  the  beam  is  consid- 
ered. 

*  And  12  feet  iu  length  between  supports. 


282  MECHANICS   OF   ENGINEEBLNG. 

EXAMPLE  3.  —  A  wrought-iron  rolled  I-beam  supported 
at  tlie  ends  is  to  be  loaded  uniformly  Fig.  235,  the  span 
being  equal  to  20  feet.  Its  cross-section,  Fig.  244,  has  a 
depth  parallel  to  the  web  of  15 
inches,  a  flange  width  of  5  inches. 
In  the  pocket  book  of  the  Trenton 
Co.  it  is  called  a  15-inch  light  I- 
beam,  weighing  150  Ibs.  per  yard, 
with  a  moment  of  inertia  =  523.  bi- 
quad.  inches  about  a  gravity  axis  perpendicular  to  the 
web  (i.e.,  when  the  web  is  vertical,  the  strongest  position) 
and  =  15  biq.  in.  about  a  gravity  axis  parallel  to  the  web 
(i.e.,  when  the  web  is  placed  horizontally). 

First  placing  the  web  vertically,  we  have  from  §  250, 

•pi  T 

W^=   Safe   load,  distributed,    =8—'.    With  #=12,000, 

fe, 

/!=523,  Z=240  inches,  e{  =  7}4  inches,  this  gives 
JF1=[8xl2,OOOx523]»[240x^]=27,893  Ibs. 

But  this  includes  the  weight  of  the  beam,  £=20  ft.  x'-fUbs. 
=1,000  Ibs.;  hence  a  distributed  load  of  26,902  Ibs.,  or  13.45 
tons  may  be  placed  on  the  beam  (secured  against  lateral 
yielding).  (The  pocket-book  referred  to  gives  13.27  tons 
as  the  safe  load.) 

Secondly,  placing  the  web  horizontal, 


of  Wl 


or  only  about  1/12  of  W^ 

EXAMPLE  4.—  Kequired  the  deflection  in  the  first  case  of 
Ex.  3.     From  §  250  the  deflection  at  middle  is 


__         =:  _L=A   Ri  L 

384    '    El,      384  '      fe,    '  El^    48  '  E    '    el 


FLEXUKE.      SAFE   LOADS.  283 


.-.  ^=0.384  in. 

EXAMPLE  5.  —  A  rectangular  beam  of  yellow  pine,  of  width 
b=4:  inches,  is  20  ft.  long,  rests  on  two  end  supports,  and  is 
to  carry  a  load  of  1,200  Ibs.  at  the  middle  ;  required  the 
proper  depth  h.  From  §  250 

P-lRI-lRW     1 

~U  ~     T  12  *  p" 

.-.  h2=6Pl-^-4:Rb.  For  variety,  use  the  inch  and  ton.  For 
this  system  of  units  P=0.60  tons,  .#'=0.50  tons  per  sq.  in., 
Z=240  inches  and  b=  4  inches. 

.-.  /r=(6x0.6x240)-K4x0.5x4)=108sq.  in.  .-.  £=10.  4  in. 

EXAMPLE  6.  —  Suppose  the  depth  in  Ex.  5  to  be  deter- 
mined by  the  condition  that  the  deflection  shall  be  =  Ysoo 
of  the  span  or  length.  We  should  then  have  from  §  250 

,       1     ,      1    PP 
500        48  Ei 

Using  the  inch  and  ton,  with  E=  1,200,000  Ibs.  per  sq.  in., 
which  =  600  tons  per  sq.  inch,  and  /=1/12fcA3,  we  have 

500X0.60X240X240X12 


As  this  is  >  10.4  the  load  would  be  safe,  as  well. 

EXAMPLE  7.  —  Eequired  the  length  of  a  wro't  iron  pipe 

supported  at  its  extremities,  its  internal  radius   being  2^ 

in.,  the  external  2.50  in.}  that  the  deflection  under  its  own 

weight  may  equal  l/m  of  the  length.  579.6  in.  Ans. 

EXAMPLE  8.—  Fig.  245.     The  wall  is  6  feet  high  and  one 

foot  thick,  of  common  brick  work 

|   i  |   i  |  i  |  i  |  i  |  i  |  [    (see  §  7)  and  is  to  be  borne  by  an 

]  i   ]  i  |  i  ]  i  ]  i  jjzj    /-beam  in  whose  outer  fibres  no 

.1  ,  =r~l    greater   normal   stress  than  8,000 

£^  ~"E  Ibs.  per  sq.  inch  is  allowable.     If 

FIG.  245.  a  number  of  I-beams  is  available, 


284  MECHANICS   OF   ENGLNEEKING. 

ranging  in  height  from  6  in.  to  15  in.  (by  whole  inches), 
which  one  shall  be  chosen  in  the  present  instance,  if  their 
cross-seetions  are  Similar  Figures,  the  moment  of  inertia  of 
the  15-inch  beam  being  800  biquad.  inches  ? 

The  12-inch  beam.  Ans. 


SHEARING  STRESSES  IN  FLEXURE. 

253.  Shearing  Stresses  in  Surfaces  Parallel  to  the  Neutral 
Surface. — If  a  pile  of  boards  (see  Fig.  246)  is  used  to  sup- 
port a  load,  the  boards  being  free  to  slip  on  each  other,  it 
;.s  noticeable  that  the  ends  overlap,  although  the  boards 


are  of  equal  length  (now  see  Fig.  247)  ;  i.e.,  slipping  has 
occurred  along  the  surfaces  of  contact,  the  combina- 
tion being  no  stronger  than  the  same  boards  side  by 
side.  If,  however,  they  are  glued  together,  piled  as  in  the 
former  figure,  the  slipping  is  prevented  and  the  deflection 
is  much  less  under  the  same  load  P.  That  is,  the  com- 
pound beam  is  both  stronger  and  stiffer  than  the  pile  of 
loose  boards,  but  the  tendency  to  slip  still  exists  and  is 
known  as  the  "  shearing  stress  in  surfaces  parallel  to  the 
neutral  surface."  Its  intensity  per  unit  of  area  will  now 
be  determined  by  the  usual  "  free-body  "  method.  In  Fig. 
248  let  AN'  be  a  portion,  considered  free,  on  the  left  of  any 


Pio.  848. 


SHEAR   IX   FLEXURE. 


285 


section  Nf,  of  a  prismatic  beam  slightly  bent  under  forces 
in  one  plane  and  perpendicular  to  the  beam.  The  moment 
equation,  about  the  neutral  axis  at  N',  gives 


•2—  =M' ;  whence  jt)'= 


M'e 


(1) 


Similarly,  with  AN  as  a  free  body,  NN1  being  =dx, 


=         w 


p  and  p'  are  the  respective  normal  stresses  in  the  outer 
fibre  in  the  transverse  sections  N  and  N'  respectively. 

Now  separate  the  block  NN',  lying  between  these  two 
consecutive  sections,  as  a  free  body  (in  Fig.  249).     And 


Pa 


furthermore  remove  a  portion  of  the  top  of  the  latter  block, 
the  portion  lying  above  a  plane  passed  parallel  to  the  neu- 
tral surface  and  at  any  distance  z"  from  that  surface.  This 
latter  free  body  is  shown  in  Fig.  250,  with  the  system  of 
forces  representing  the  actions  upon  it  of  the  portions  taken 
away.  The  under  surface,  just  laid  bare,  is  a  portion  of  a  sur- 
face (parallel  to  the  neutral  surface)  in  which  the  above  men- 
tioned slipping,  or  shearing,  tendency  exists.  The  lower  por- 
tion (of  the  block  NN'}  which  is  now  removed  exerted  this 


280  MECHANICS   OF   ENGINEERING. 

rubbing,  or  sliding,  force  on  the  remainder  along  the  under 
surface  of  the  latter.  Let  the  unknown  intensity  of  this 
shearing  force  be  JT(per  unit  of  area)  ;  then  the  shearing 
force  on  this  under  surface  is  =Xy"dx,  (y",=  oa  in  figure, 
being  the  horizontal  width  of  the  beam  at  this  distance  z" 
from  the  neutral  axis  of  N')  and  takes  its  place  with  the 
other  forces  of  the  system,  which  are  the  normal  stresses 

between    '  ,    and  portions  of  J  and  J',  the  respective 

\_z=  z" 

total  vertical  shears.  (The  manner  of  distribution  of  J 
over  the  vertical  section  is  as  yet  unknown  ;  see  next  arti- 
cle.) 

Putting  S  (horiz.  compons.)  =  0  in  Fig.  250,  we  have 

fe  -p'dF—  Ce  Z-pdF—Xy"dx=Q 

Jz"     e  Jz"     e 

.-.Xy"dx=£—P      fldF 
e  z" 

But  from  eqs.  (1)  and  (2),  /—  p  =  (M'—M)±.=^r  dM, 
while  from  §  240  dM  =  Jdx  ; 


....    (3) 

as  the  required  intensity  per  unit  of  area  of  the  shearing 
force  in  a  surface  parallel  to  the  neutral  surface  and  at  a 
distance  z"  from  it.  It  is  seen  to  depend  on  the  "  shear  "  J 
and  the  moment  of  inertia  I  of  the  whole  vertical  section; 
upon  the  horizontal  thickness  y"  of  the  beam  at  the  sur- 

face   in    question  ;    and    upon     the     integral      J      zdF, 

which  (from  §  23)  is  the  product  of  the  area  of  that  part  of 
the  vertical  section  extending  from  the  surface  in  question  to 
the  outer  fibre,  by  the  distance  of  the  centre  of  gravity  of  that 
part  from  the  neutral  surface. 


SHEAR   IX    FLEXURE. 


287 


It  now  follows,  from  §  209,  that  the  intensity  (per  unit 
area)  of  the  shear  on  an  elementary  area  of  the  vertical 
cross  section  of  a  bent  beam,  and  this  intensity  we  may  call 
Z,  is  equal  to  that  X,  just  found,  in  the  horizontal  section 
which  is  at  the  same  distance  (z")  from  the  neutral  axis. 

254.  Mode  of  Distribution  of  J,  the  Total  Shear,  over  the  Verti- 
cal Cross  Section. — The  intensity  of  this  shear,  Z  (Ibs.  per 
sq.  inch,  for  instance)  has  just  been  proved  to  be 


=X=~^,     C 

l     J' 


zdF 


(4) 


To  illustrate  this,  required  the 
value  of  Z  two  inches  above  the  neu- 
tral axis,  in  a  cross  section  close  to 
the  abutment,  in  Ex.  5,  §  252.  Fig. 
251  shows  this  section.  From  it  we  AS 
have  for  the  shaded  portion,  lying 
above  the  locality  in  question,  y"  = 


4  inches,  and 


=5.2 


r 

j  2//= 


zdF  =  (area 


of  shaded  portion)  X  (distance  of 
its  centre  of  gravity  from  2JA)  = 
(12.8  sq.  in.)  x  (3.6  in.)  =  46.08  cubic  inches. 

The  total  shear  J  =  the  abutment  reaction  =  600  Ibs., 
while  /  =  L  W  =  £.  x  4  x  (10.4)3  =  375  biquad.  inches. 
Both  t/and  /refer  to  the  whole  section. 

7    600x46.08    -.Q.o!, 
.'.Z=  —  ._  —  -  —  =18.42  Ibs.  per  sq.  in., 
d75  x4 

qui+e  insignificant.     In  the  neighborhood  of  the  neutral 
axis,  where  z"  =  0,  we  have  y"  =  4  and 


Ce        zdF=  rezdF  =  20.8  x  2.6=54.8, 
J  z"=0  J  0 

wh__e  J  and  /  of  course  are  the  same  as  before.     Hence 
for  z"  =0 


^ 


2SS  MECHANICS   OF   ENGINEERING. 

Z=Z0=  21.62  Ibs.  per  sq.  in. 

At  the  outer  fibre  since  /*  *dF=0,  z"  being  =  e,  Z  is  =  0 
«/  e 

for  a  beam  of  any  shape. 

For  a  solid  rectangular  section  like  the 
above,  Z  and  z '  bear  the  same  relation  to 
each  other  as  the  co-ordinates  of  the  para- 
bola in  Fig.  252  (axis  horizontal). 

Since  in  equation  (4)  the  horizontal 
thickness,  y",  from  side  to  side  ef  the  sec- 
tion of  the  locality  where  Z  is  desired, 

occurs  in  the  denominator,  and  since   /    zdF 

t/,// 

increases  as  z'1  grows  numerically  smaller,  the  following 
may  be  stated,  as  to  the  distribution  of  «7,  the  shear,  in 
any  vertical  section,  viz.: 

The  intensity  (Ibs.  per  sq.  in.)  of  the  shear  is  zero  at 
the  outer  elements  of  the  section,  and  for  beams  of  ordi- 
nary shapes  is  greatest  where  the  section  crosses  the  neu- 
tral surface.  For  forms  of  cross  section  having  thin  webs 
its  value  may  be  so  great  as  to  require  special  investiga- 
tion for  safe  design. 

Denoting  by  ZQ  the  value  of  Z  at  the  neutral  axis,  (which 
= X0  in  the  neutral  surface  where  it  crosses  the  vertica 
section  in  question)  and  putting  the  thickness  of  the  sub- 
stance of  the  beam  =  10  at  the  neutral  axis,  we  have, 

T        (  area    above  )       (  ,,      -,.   ,      , .,  ,    ) 

Z0=X0=-J-  X  1  neutralaxis  t  x  J  thedlst-  °fi*soent.  (  (6) 
H>«       (  (or  below)  f      (  grav.from  that  axis  p 

255.  Values  of  Zo  for  Special  Forms  of  Cross  Section. — From 
the  last  equation  it  is  plain  that  for  a  prismatic  beam  the 
value  of  Z0  is  proportional  to  J,  the  total  shear,  and  hence 
to  the  ordinate  of  the  shear  diagram  for  any  particular 
case  of  loading.  The  utility  of  such  a  diagram,  as  obtain 


SHEAB  IK   FLEXURE. 


289 


ed  in  Figs.  234-237  inclusive,  is  therefore  evident,  for  by 
locating  the  greatest  shearing  stress  in  the  beam  it 
enables  us  to  provide  proper  relations  between  the  load- 
ing and  the  form  and  material  of  the  beam  to  secure  safety 
against  rupture  by  shearing. 

The  table  in  §  210  gives  safe  values  which  the 
maximum  Z0  in  any  case  should  not  exceed.  It  is 
only  in  the  case  of  beams  with  thin  webs  (see  Figs. 
238  and  240)  however,  that  ZQ  is  likely  to  need  at- 
tention. 

For  a  Rectangle  we  have,  Fig.  253,  (see  eq.  5,  § 


254)  b0=b,  I=Y,26AS,  and. 


.:Z0=X0=-^-  2.  i.e.,  =  |-  (total  shear)-;-  (whole  area) 

Hence  the  greatest  intensity  of  shear  in  the  cross-section 
is  A  as  great  per  unit  of  area  as  if  the  total  shear  were 
uniformly  distributed  over  the  section. 


FIG.  254.  FIG.  355.  FIG.  256. 

"For  a  Solid  Circular  section  Fig.  254 


7=  J  re^F= 

Ib0J° 


.  2r      2      3,T     3 


[See  §  26  Prob.  3]. 

For  a  Hollow  Circular  section  (concentric    circles)    Fig. 
255,  we  have  similarly, 


290 


MECHAiaCS  OF  ENGINEERING. 

J  VTIT       frxr 


1-r2)     2     ' 


2 


3 


Applying  this  formula  to  Example  2  §  252,  we  first  have 
as  the  max.  shear  «7m  =  */2P  =1,735  Ibs.,  this  being  the  abut- 
ment reaction,  and  hence  (putting  x  =  (22  -r-  7)) 


which  cast  iron  is  abundantly  able  to  withstand  in  shear- 
ing. 

For   a  Hollow  Rectangular  Beam,  symmetrical   about    its 
neutral  surface,  Fig.  256  (box  girder) 


The  same  equation  holds  good  for  Fig.  257  (I-beam  with 
square  corners)  but  then  62  denotes  the  sum  of  the  widths 
of  the  hollow  spaces. 

256.  Shearing  Stress  in  the  Web  of  an  I-Beam. —  It  is  usual  to 
consider  that,  with  I-beams  (and  box- 
beams)  with  the  web  vertical  the  shear  J, 
in  any  vertical  section,  is  borne  exclusively 
by  the  web  and  is  uniformly  distributed 
over  its  section.  That  this  is  nearly  true 
may  be  proved  as  follows,  the  flange  area 
being  comparatively  large.  Fig.  258.  Let 
FI  be  the  area  of  one  flange,  and  F0  that  of 
the  half  web.  Then  since 

FIG.  258. 


SHEAR  IX  FLEXURE.  291 

(the  last  term  approximate,  ^  J^  being  taken  for  the  radi- 
us of  gyration  of  F},}  while 


F=^V— +^o-j-,   (the  first  term  approx.)  we  have 

ff 

rr  **  /4-  "VV**-*-   1    I    -*•  U/ 

^o== 7f =  i  M  ?. /a  y  .1.0  *M  »  which  = 


if  we  write  (2^+^,)  -4-  (6^+2^)  =K  •  But  Mo  is  the 
area  of  the  whole  web,  .'.  the  shear  per  unit  area  at  the 
neutral  axis  is  nearly  the  same  as  if  J  were  uniformly  dis- 
tributed over  the  web.  E.  g.,  with  FI  =  2  sq.  in.,  and  FQ 
=  1  sq.  in.  we  obtain  Z0  =  1.07  (J-t-baho). 

Similarly,  the   shearing  stress  per   unit   area  at  n,  the 
upper  edge  of  the  web,  is  also  nearly  equal  to  J  -f-  b^  (see 


eq.,  4  (§254)  for  then   \T   (zdF)l=     F^/h,       nearly, 

while  /  remains  as  before. 

The  shear  per  unit  area,  then,  in  an  ordinary  I-beam  is 
obtained  by  dividing  the  total  shear  J  by  the  area  of  the 
web  section. 

EXAMPLE.—  It  is  required  to  determine  the  proper 
thickness  to  be  given  to  the  web  of  the  15-inch  wrought- 
iron  rolled  beam  of  Example  3  of  §252,  the  height  of  web 
being  13  inches,  with  a  safe  shearing  stress  as  low  as  4000 
Ibs.  per  sq.  in.  (the  practice  of  the  N.  J.  Steel  and  Iron 
Co.,  for  webs),  the  web  being  vertical. 

The  greatest  total  shear,  Jmt  occurring  at  either  support 
and  being  equal  to  half  the  load  (see  table  §250)  we  have 
with  60  =  width  of  web, 


Z0  max.=       -;  i.e.  4000  =  ...  ^  =  0.26  inches. 


292 


MECHANICS   OF   ENGINEERING. 


(Units,  inch  and  pound).  The  15-inch  light  beam  of  the 
N.  J.  Co.  has  a  web  ]/%  inch  thick,  so  as  to  provide  for  a 
shear  double  the  value  of  that  in  the  foregoing  example. 
In  the  middle  of  the  span  Z^  =  0,  since  J  =  0. 

257.  Designing  of  Riveting  for  Built  Beams. — The  latter  are 
generally  of  the  I-beam  and  box  forms,  made  by  riveting 
together  a  number  of  continuous  shapes,  most  of  the  ma- 
terial being  thrown  into  the  flange  members.  E.  g.,  in  fig. 
259,  an  I-beam  is  formed  by  riveting  together,  in  the 
manner  shown  in  the  figure,  a  "vertical  stem  plate"  or 
web,  four  "  angle-irons,"  and  two  "  flange-plates,"  each  of 


these  seven  pieces  being  continuous  through  the  whole 
length  of  the  beam.  Fig  260  shows  a  box-girder.  If  the 
riveting  is  well  done,  the  combination  forms  a  single  rigid 
beam  whose  safe  load  for  a  given  span  may  be  found  by 
foregoing  rules  ;  in  computing  the  moment  of  inertia,  how- 
ever, the  portion  of  cross  section  cut  out  by  the  rivet 
holes  must  not  be  included.  (This  will  be  illustrated  in 
a  subsequent  paragraph.)  The  safe  load  having  been  com- 
puted from  a  consideration  of  normal  stresses  only,  and 
the  web  being  made  thick  enough  to  take  up  the  max, 
total  shear,  Jm,  with  safety,  it  still  remains  to  design  the 
riveting,  through  whose  agency  the  web  and  flanges  are 
caused  to  act  together  as  a  single  continuous  rigid  mass. 
It  will  be  on  the  side  of  safety  to  consider  that  at  a  given 


SHEAR  IN   FLEXURE.  293 

locality  in  the  beam  the  shear  carried  by  the  rivets  con- 
necting the  angles  and  flanges,  per  unit  of  length  of  beam, 
is  the  same  as  that  carried  by  those  connecting  the  angles 
and  the  web  ("vertical  stem-plate").  The  amount  of  this 
shear  may  be  computed  from  the  fact  that  it  is  equal  to 
that  occurring  in  the  surface  (parallel  to  the  neutral  sur- 
face) in  which  the  web  joins  the  flange,  in  case  the  web 
and  flange  were  of  continuous  substance,  as  in  a  solid  I- 
beam.  But  this  shear  must  be  of  the  same  amount  per 
horizontal  unit  of  length  as  it  is  per  vertical  linear  unit  in 
the  web  itself,  where  it  joins  the  flange  ;  (for  from  §  254  Z 
=X.)  But  the  shear  in  the  vertical  section  of  the  web, 
being  uniformly  distributed,  is  the  same  per  vertical  linear 
unit  at  the  junction  with  the  flange  as  at  any  other  part 
of  the  web.  section  (§  256,)  and  the  whole  shear  on  the  ver- 
tical section  of  web  =  J,  the  "  total  shear  "  of  that  section 
of  the  beam. 

Hence  we  may  state  the  following : 

The  riveting  connecting  the  angles  with  the  flanges,  (or 
the  web  with  the  angles)  in  any  locality  of  a  built  beam, 
must  safely  sustain  a  shear  equal  to  Jon  a  horizontal  length 
equal  to  the  height  of  iveb. 

The  strength  of  the  riveting  may  be  limited  by  the  re- 
sistance of  the  rivet  to  being  sheared  (and  this  brings 
into  account  its  cross  section)  or  upon  the  crushing  resist- 
ance of  the  side  of  the  rivet  hole  in  the  plate  (and  this  in- 
volves both  the  diameter  of  the  rivet  and  the  thickness  of 
the  metal  in  the  web,  flange,  or  angle.)  In  its  practice  the 
N.  J.  Steel  and  Iron  Co.  allows  7500  Ibs.  per  sq.  inch  shear- 
ing stress  in  the  rivet  (wrought  iron),  and  12500  Ibs.  per 
sq.  inch  compressive  resistance  in  the  side  of  the  rivet- 
hole,  the  axial  plane  section  of  the  hole  being  the  area  of 
reference. 

In  fig.  259  the  rivets  connecting  the  web  with  the  angles 
are  in  double  shear,  which  should  be  taken  into  account  in 
considering  their  shearing  strength,  which  is  then  double  ; 
those  connecting  the  angles  and  the  flange  plates  are  in 


294  MECHANICS   OF   ENGINEERING. 


In  fig.  260  (box -beam)  where  the  beam  is 
built  of  two  webs,  four  angles,  and  two  flange  plates,  all 
the  rivets  are  in  single  shear.  If  the  web  plate  is  very 
high  compared  with  its  thickness,  vertical  stiffeners  in  the 
form  of  T  irons  may  need  to  be  riveted  upon  them  lat- 
erally [see  §  314]. 

EXAMPLE. — A  built  I-beam  of  wrought  iron  (see  fig.  259) 
is  to  support  a  uniformly  distributed  load  of  40  tons,  its 
extremities  resting  on  supports  20  feet  apart,  and  the 
height  and  thickness  of  web  being  20  ins.  and  y2  in.  re- 
spectively. How  shall  the  rivets,  which  are  y  in.  in  di- 
ameter, be  spaced,  between  the  web  and  the  angles  which 
are  also  y2  in.  in  thickness  ?  Keferring  to  fig.  235  we  find 
that  J  =  y£  W  =  20  tons  at  each  support  and  diminishes 
regularly  to  zero  at  the  middle,  where  no  riveting  will  there- 
fore be  required.  (Units  inch  and  pound).  Near  a  sup- 
port the  riveting  must  sustain  for  each  inch  of  length  of 
beam  a  shearing  force  of  (J  -r-  height  of  web)  =  40000  -r- 
20  in.  =  2000  Ibs.  Each  rivet,  having  a  sectional  area  of 
y^  7i  (7/%f  =  0.60  sq.  inches,  can  bear  a  safe  shear  of  0.60 
X  7500  =  4500  Ibs.  in  single  shear,  and  .-.  of  9000  Ibs.  in 
double  shear,  which  is  the  present  case.  But  the  safe 
compressive  resistance  of  the  side  of  the  rivet  hole  in 
either  the  web  or  the  angle  is  only  ^  in.  x  l/i  in.  x  12500 
=  5470  Ibs.,  and  thus  determines  the  spacing  of  the  rivets 
as  follows  : 

2000  Ibs.  -r-  5470  gives  0.36  as  the  number  of  rivets  per 
inch  of  length  of  beam,  i.e.,  they  must  be  1  4-  0.36  =  2.7 
inches  apart,  centre  to  centre,  near  the  supports;  5.4  inches 
apart  at  ^  the  span  from  a  support ;  none  at  all  in  the 
middle. 

However,  "  the  rivets  should  not  ba  spaced  closer  than 
2^  times  their  diameter,  nor  farther  apart  than  16  times 
the  thickness  of  the  plate  they  connect,"  is  the  rule  of  the 
N.  J.  Co. 

As  for  the  rivets  connecting  the  angles  and  flange  plates, 
being  in  two  rows  and  opposite  (in  pairs)  the  safe  shear- 


FLEXUKE.  BUILT   BEAM. 


295 


ing  resistance  of  a  pair  (each  in  single  shear)  is  9,000  Ibs., 
while  the  safe  compressive  resistance  of  the  sides  of  the 
two  rivet  holes  in  the  angle  irons  (the  flange  plate  being 
much  thicker)  is  —  10,940  Ibs.  Hence  the  former  figure 
(9,000)  divided  into  2,000  Ibs.,  gives  0.22  as  the  number  of 
pairs  of  rivets  per  inch  of  length  of  the  beam  ;  i.e.,  the 
rivets  in  one  row  should  be  spaced  4.5  inches  apart,  centre 
to  centre,  near  a  support ;  the  interval  to  be  increased  in 
inverse  ratio  to  the  distance  from  the  middle  of  span, 
^bearing  in  mind  the  practical  limitation  just  given). 

If  the  load  is  concentrated  in  the  middle  of  the  span, 
instead  of  uniformly  distributed,  «7is  constant  along  each 
half-span,  (see  fig.  234)  and  the  rivet  spacing  must  accord- 
ingly be  made  the  same  at  all  localities  of  the  beam. 


SPECIAL,  PROBLEMS  IN  FLEXURE. 


258.  Designing  Cross  Sections  of  Built  Beams. — The  last  par- 
agraph dealt  with  the  riveting  of  the  various  plates ;  we 
now  consider  the  design  of  the  plates  themselves.  Take 
for  instance  a  built  I-beam,  fig.  261 ;  one  vertical  stem- 


FIQ.  261. 


296  MECHANICS   OF   ENGINEERING, 

plate,  four  angle  irons,  (each  of  sectional  area  =  A,  re- 
maining after  the  holes  are  punched,  with  a  gravity  axis 
parallel  to,  and  at  a  distance  =  a  from  its  base),  and  two 
flange  plates  of  width  =  b,  and  thickness  =  t.  Let  the 
whole  depth  of  girder  =  h,  and  the  diameter  of  a  rivet 
hole  =1f.  To  safely  resist  the  tensile  and  compressive 
forces  induced  in  this  section  by  Mm  inch-lbs.  (Mm  being 
the  greatest  moment  in  the  beam  which  is  prismatic)  we 
have  from  §  239, 

*.=  £?  a) 

E1  for  wrought  iron  =  12,000  Ibs.  per  sq.  inch,  e  is  =  */2  h 
while  1,  the  moment  of  inertia  of  the  compound  section, 
is  obtained  as  follows,  taking  into  account  the  fact  that 
the  rivet  holes  cut  out  part  of  the  material.  In  dealing 
with  the  sections  of  the  angles  and  flanges,  we  consider 
them  concentrated  at  their  centres  of  gravity  (an  approx- 
imation, of  course,)  and  treat  their  moments  of  inertia 

about  N  as  single  terms  in  the  series    CdF  z* 

(see  §  85).  The  subtractive  moments  of  inertia  for  the 
rivet  holes  in  the  web  are  similarly  expressed  ;  let  b0  = 
thickness  of  web. 

7N  for  web  =   -Jyb0  (h— 2£)3— 26/  [} — t— a']2 
7N  for  four  angles  =  4 A  [~ — t — a]2 
7N  for  two  flanges  =  2(6— 2f)  t  (1£)2 

the  sum  of  which  makes  the  1^  of  the  girder.  Eq.  (1)  may 
now  be  written 

-^=7*  (2) 


which  is  available  for  computing  any  one  unknown  quan- 
tity. The  quantities  concerned  in  /N  are  so  numerous  and 
they  are  combined  in  so  complex  a  manner  that  in  any 
numerical  example  it  is  best  to  adjust  the  dimensions  of 
the  section  to  each  other  by  successive  assumptions  and 


FLEXURE   BUILT   BEAM.  297 

trials.  The  size  of  rivets  need  not  vary  much  in  different 
cases,  nor  the  thickness  of  the  web-plate,  which  as  used 
by  the  N.  J.  Co.  is  "  rarely  less  than  ^  or  more  than  ^ 
inch  thick."  The  same  Co.  recommends  the  use  of  a 
single  size  of  angle  irons,  viz.,  3"  X  3''  X  ^",  for  built 
girders  of  heights  ranging  from  12  to  36  inches,  and  also 
y±  in.  rivets,  and  gives  tables  computed  from  eq.  (2)  for 
the  proportionate  strength  of  each  portion  of  the  com- 
pound section. 

EXAMPLE. — (Units,  inch  and  pound).  A  built  I-beam 
with  end  supports,  of  span  =  20  ft.  =  240  inches,  is  to 
support  a  uniformly  distributed  load  of  36  tons  =  72,000  Ibs. 
If  y±  inch  rivets  are  used,  angle  irons  3''  x  3"  X  y2",  ver- 
tical web  ^4"  in  thickness,  and  plates  1  inch  thick  for 
flanges,  how  wide  (b  =  ?)  must  these  flange-plates  be  ? 
taking  h  =  22  inches  =  total  height  of  girder. 

Solution.— From  the  table  in  §  250  we  find  that  the  max. 
M  for  this  case  is  *£  Wl,  where  W  =  the  total  distributed 
load  (including  the  weight  of  the  girder)  and  I  =  span. 
Hence  the  left  hand  member  of  eq.  (2)  reduces  to 

Wl     h_        72000  x  240  x  22 
16  '  X"  ~          16  x  12000 

That  is,  the  total  moment  of  inertia  of  the  section  must 
be  =  1,980  biquad.  inches,  of  which  the  web  and  angles 
supply  a  known  amount,  since  b0  =  *£",  t  =1",  <*=  ^", 
a'  =  1%",  A=  2.0  sq.  in.,  a  =  0.9',  and  h  =  22",  are 
known,  while  the  remainder  must  be  furnished  by  the 
flanges,  thus  determining  their  width  b,  the  unknown 
quantity. 

The  effective  area,  A,  of  an  angle  iron  is  found  thns : 
The  full  sectional  area  for  the  size  given,  =  3  x  */4  + 
2^  X  YZ  =2.75  sq.  inches,  from  which  deducting  for  two 
rivet  holes  we  have 

A=  2.75 — 2  x  Y±  X  #=  2.0  sq.  in. 

The  value  a  =  0.90"  is  found  by  cutting  out  the  shape 


298  MECHANICS    OF   ENGINEERING. 

of  two  angles  from  sheet  iron,  thus  : 
and  balancing  it  on  a  knife  edge.  (The 
gaps  left  by  the  rivet  holps  maybe  ignored, 
without  great  error,  in  finding  a).  Hence, 
substituting  we  have 


Iv  for  web  =i 
INfor  four  angles  =  4x2x  [9.10]2=662.5 
IN  for  two  flanges=2(^7)xlX(10X)2=220.4(&—  1.5) 

.-.  1980=  282.3+662.5+(fr—  1.5)220.4 
whence  b  =  4.6  +  1.5  =  6.1  inches 

the  required  total  width  of  each  of  the  1  in.  flange  plates. 
This  might  be  increased  to  6.5  in.  so  as  to  equal  the 
united  width  of  the  two  angles  and  web. 

The  rivet  spacing  can  now  be  designed  by  §  257,  and 
the  assamed  thickness  of  web,  ^4  in.,  tested  for  the  max. 
total  shear  by  §  256.  The  latter  test  results  as  follows  : 
The  max.  shear  Jm  occurs  near  either  support  and  = 
y2  W=  36,000  Ibs.  .-.,  calling  Z/o  the  least  allowable  thickness 
of  web  in  order  to  keep  the  shearing  stress  as  low  as  4,000 
Ibs.  per  sq.  inch, 

b'0x  20"  x  4000  =36000  .-.  6'0=0.45" 

showing  that  the  assumed  width  of  ^  in.  is  safe. 

This  girder  will  need  vertical  stiffeners  near  the  ends, 
as  explained  subsequently,  and  is  understood  to  be  sup- 
ported laterally.  Built  beams  of  double  web,  or  box- 
form,  (see  Fig.  260)  do  not  need  this  lateral  support. 

259.  Set  of  Moving  Loads.  —  "When  a  locomotive  passes  over 
a  number  of  parallel  prismatic  girders,  each  one  of  which 
experiences  certain  detached  pressures  corresponding  to 
the  different  wheels,  by  selecting  any  definite  position  of 
the  wheels  on  the  span,  we  may  easily  compute  the  reac- 
tions of  the  supports,  then  form  the  shear  diagram,  and 
finally  as  in  §  243  obtain  the  max.  moment,  Mm)  and  the 


FLEXUKE.       MOVING   LOADS. 


1>99 


max.  shear  Jm,  for  this  particular  position  of  the  wheels. 
But  the  values  of  Mm  and  Jm  for  some  other  position  may 
be  greater  than  those  just  found.  We  therefore  inquire 
which  will  be  the  greatest  moment  among  the  infinite 
number  of  (Jfm)'s  (one  for  each  possible  position  of  the 
wheels  on  the  span).  It  is  evident  from  Fig.  236  from  the 
nature  of  the  moment  diagram,  that  when  the  pressures  or 
loads  are  detached,  the  Mm  for  any  position  of  the  loads, 
which  of  course  are  in  this  case  at  fixed  distances  apart, 
must  occur  under  one  of  the  loads  (i.e.  under  a  wheel). 
We  begin  /.  by  asking :  What  is  the  position  of  the  set  of 
moving  loads  when  the  moment  under  a  given  wheel  is 
greater  than  will  occur  under  that  wheel  in  any  other  po- 
sition ?  For  example,  in  Fig.  262,  in  what  position  of  the 


a* 


.-i t=L_MJ.,fr_ 

O  ffi 1 00 


r 


=r~i 

p« 

j 


loads  PI,  P2,  etc.  on  the  span  will  the  moment  M2,  i.e., 
under  P2,  be  a  maximum  as  compared  with  its  value  under 
P2  in  any  other  position  on  the  span.  Let  R  be  the  resultant 
of  the  loads  which  are  now  on  the  span,  its  variable  distance 
from  0  be  =  x,  and  its  fixed  distance  from  P2  =  a'\  while 
a,  b,  c,  etc.,  are  the  fixed  distances  between  the  loads 
(wheels).  For  any  values  of  x  ,  as  the  loading  moves 
through  the  range  of  motion  within  which  no  wheel  of  the 
set  under  consideration  goes  off  the  span,  and  no  new- 
wheel  comes  on  it,  we  have  S1  =--^-  R,  and  the  moment 
under  P2 


(1) 


300  MECHANICS   OF   ENGINEERING. 

In  (1)  we  have  M.2  as  a  function  of  x,  all  the  other  quan- 
tities in  the  right  hand  member  remaining  constant  as  the 
loading  moves;  x  may  vary  from  x=  a+a'  to 
x—l—  (c+&—  a').     For  a  max.  M2,  we  put  dM2^-dx=Q,  i.  e. 

~(l-2x~+a')=0  .-.  x  (for  Max  M,}=y2l+y2a' 


(For  this,  or  any  other  value  of  x,  cPM.^dx2  is  negative, 
hence  a  maximum  is  indicated).  For  a  max.  M2,  then,  R 
must  be  as  as  far  (y2af)  on  one  side  of  the  middle  of  the 
span  as  P2  is  on  the  other  ;  i.e.,  as  the  loading  moves,  the 
moment  under  a  given  wheel  becomes  a  max.  when  that 
wheel  and  the  centre  of  gravity  of  all  the  loads  (then  on 
the  span)  are  eg  ui  -distant  from  the  middle  of  the  span. 

In  this  way  in  any  particular  case  we  may  find  the- 
respective  max.  moments  occurring  under  each  of  the 
wheels  during  the  passage,  and  the  greatest  of  these  is  the 
Mm  to  be  used  in  the  equation  M^^—R'I-^e  for  safe  loading.* 

As  to  the  shear  J,  for  a  given  position  of  the  wheels  this 
will  be  the  greatest  at  one  or  the  other  support,  and 
equals  the  reaction  at  that  support.  When  the  load  moves 
toward  either  support  the  shear  at  that  end  of  the  beam 
evidently  increases  so  long  as  no  wheel  rolls  completely 
over  and  beyond  it.  To  find  «/max.,  then,  dealing  with 
each  support  in  turn,  we  compute  the  successive  reactions 
at  the  support  when  the  loading  is  successively  so  placed 
that  consecutive  wheels,  in  turn,  are  on  the  point  of  roll- 
ing off  the  girder  at  that  end  ;  the  greatest  of  these  is  the 
max.  shear,  Jm.  As  the  max.  moment  is  apt  to  come  under 
the  heaviest  load  it  may  not  be  necessary  to  deal  with 
more  than  one  or  two  wheels  in  finding  Mm. 

EXAMPLE.  —  Given  the  following  wheel  pressures, 

A<  .  .  &  .  .  >B<  .  .  5'  .  .  >C<  .  .  4  .  .  <D 
4  tons.  6  tons.  6  tons.  5  tons. 

on  one  rail  which  is  continuous  over  a  girder  of  20  ft.  span, 
under  a  locomotive. 

*  Since  this  may  be  regarded  as  a  case  of  "  sudden  application"  of  a  load,  it  is 
customary  to  make  R'  much  smaller  than  for  a  dead  load;  from  one-third  to  one-half 
smaller. 


FLEXURE.      MOVING  LOADS. 


301 


1.  Eequired  the  position  of  the  resultant  of  A,  B,  and  C  > 

2.  "  "  "  "        A,  B,  C,  and  D ; 

3.  "  "  "  "  B,  (7,  and  D. 

4.  In  what  position  of  the  wheels  on  the  span  will  the 
moment  under  B  be  a  max.  ?    Ditto  for  wheel  C?    Eequired 
the  value  of  these  moments  and  which  is  Mm  ? 

5.  Eequired  the  value  of  Jm,  (max.  shear),  its  location  and 
the  position  of  loads. 

Results.— (1.)  T.ff  to  right  of  A.  (2.)  10*  to  right  of  A. 
(3.)  4.4'  to  right  of  B.  (4.)  Max.  MB  »=  1,273,000  inch  Ibs. 
with  all  the  wheels  on  ;  Max.  Mc  =  1,440,000  inch-lbs.  with 
wheels  B,  C,  and  D  on.  (5.)  Jm  =  13.6  tons  at  right  sup- 
port with  wheel  D  close  to  this  support. 

260.  Single  Eccentric  Load. — In  the  following  special  cases 
of  prismatic  beams,  peculiar  in  the  distribution  of  the 
loads,  or  mode  of  support,  or  both, 
the  main  objects  sought  are  the 
values  of  the  max.  moment  Mm)  for 
use  in  the  equation 

J4=^(see  §239); 

and  of  the  max.  shear  Jm,  from 
which  to  design  the  web  riveting 
in  the  case  of  an  /  or  box-girder. 
The  modes  of  support  will  be  such 
that  the  reactions  are  independent 
of  the  form  and  material  of  the 
beam  (the  weight  of  beam  being 
neglected).  As  before,  the  flexure  is  to  be  slight,  and  the 
forces  are  all  perpendicular  to  the  beam. 

The  present  problem  is  that  in  fig.  263,  the  beam  being 
prismatic,  supported  at  the  ends,  with  a  single  eccentric 
load,  P.  We  shall  first  disregard  the  weight  of  the  beam 
itself.  Let  the  span  =  ?!+£,.  First  considering  the  whole 
beam  free  we  have  the  reactions  72,  =  P^  -f-  I  and  R2  — 
PI,  -  Z. 

Making  a  section  at  m  and  having  Om  free,  x  being  <  12, 
-T  (vert,  compons.)  =  0  gives 


FIG.  sea. 


302  MECHANICS   OF   ENGINEERING. 

R,  —  e/=0,  Le.,  J=R2  ; 
while  from  2  (mom.)m=0  we  have 

P-Z-R,x=  0  .\M=  fi&=S& 

e  I 

These  values  of  J  and  M  hold  good  between  0  and  (7,  «/ 
being  constant,  while  M  is  proportional  to  x.  Hence  for 
0  G  the  shear  diagram  is  a  rectangle  and  the  moment  dia- 
gram a  triangle.  By  inspection  the  greatest  M  for  00  is 
for  x  —  Z2,  and  =  Pl^  -r-  1.  This  is  the  max.  M  for  the 
beam,  since  between  G  and  B,  M  is  proportional  to  the  dis- 
tance of  the  section  from  B. 


is  the  equation  for  safe  loading. 

J  —  Rl  in  any  section  along  CB,  and  is  opposite  in  sign 
to  what  it  is  on  0(7;  i.e.,  practically,  if  a  dove-tail  joint 
existed  anywhere  on  OC  the  portion  of  the  beam  on  the 
right  of  such  section  would  slide  downward  relatively  to 
the  left  hand  portion  ;  but  vice  versa  on  CB. 

Evidently  the  max.  shear  Jm  =  fy  or  R2,  as  12  or  Z:  is  the 
greater  segment. 

It  is  also  evident  that  for  a  given  span  and  given  beam 
the  safe  load  P',  as  computed  from  eq.  (1)  above,  becomes 
very  large  as  its  point  of  application  approaches  a  sup- 
port ;  this  would  naturally  be  expected  but  not  without 
limit,  as  the  shear  for  sections  between  the  load  and  the 
support  is  equal  to  the  reaction  at  the  near  support  and 
may  thus  soon  reach  a  limiting  value,  when  the  safety  of 
the  web  or  the  spacing  of  the  rivets,  if  any,  is  considered. 

Secondly,  considering  the  weight  of  the  beam,  or  any 
uniformly  distributed  loading,  weighing  w  Ibs.  per  unit  of 
length  of  beam,  in  addition  to  P,  Fig.  264,  we  have  the 
reactions  • 

P     P12.  W        ,    „     PI,  ,  W 
^=T+T;and7?2=T+lT 

Let  12  be  >ll;   then  for  a  portion   Om  of   length   x<l^t 
moments  about  m  give 


FLEXURE.      SPECIAL   PROBLEMS. 


303 


. 

i.e.,  on  00,  M=K2x  -  y2  wx*  .... 

Evidently  for  x  =  0  (i.e.  at  0)  M  =  0,  while  for  x  = 
at  C)  we  have,  putting  w  =  W  -r-  I 


(2) 
(i.e. 


w, 

(  max.  ) 

-=0  .-.  xa  •<  for  M  or     >•  =- 

(  min.   ) 


(3) 
»          4  & 

It  remains  to  be  seen  whether  a  value  of  M  may  not  exist 
in  some  section  between  0  and  C,  (i.e.,  for  a  value  of  x 
<l,  in  eq.  (2)),  still  greater  than  Mc.  Since  (2)  gives  M  as 
a  continuous  function  of  x  between  0  and  C,  we  put 
d M  -r-  cfo  =  0,  and  obtain,  substituting  the  value  of  the  con- 
stants R2  and  w, 

max. 

for  Jtf  or     \  =^+  ^  I         (4) 
min. 

This  must  be  for   M  max.,   since  cPM  -*-  dx*  is  negative 

when  this  value  of  x  is  sub- 
stituted.     If   the  particular 

T  value  of  x  given  by  (4)  is 
<^,  the  corresponding  value 
of  M  (call  it  Jfn)  from  eq. 
(2)  will  occur  on  0(7  and  will 
be  greater  than  MG  (Dia- 
grams II.  in  fig.  264  show 
this  case) ;  but  if  xn  is>  l& 
we  are  not  concerned  with 

R]  the  corresponding  value  of 
M,  and  the  greatest  M  on  OC 
would  be  Mc. 

For  the  short  portion  BC, 
which  has  moment  and  shear 
diagrams  of  its  own  not  con- 
tinuous with  those  for  0(7,  it 
may  easily  be  shown  that 
MG  is  the  greatest  moment  of 
PIG.  864.  anJ  section.  Hence  the  M 


304  MECHANICS   OF   ENGINEERING. 

max.,  or  Mm)  of  the  whole  beam  is  either  Mc  or  Mn> 
according  as  xn  is  >  or  <  lz.  This  latter  criterion  may  be 
expressed  thus,  [with  12  —  y2  I  denoted  by  13,  the  distance 
of  P  from  the  middle  of  the  span]  : 


(eq. 
and  since  from  (4)  and  (2) 


The  equation  for  safe  loading  is 


and  ....     (6) 

I  j  Se^eqs  (3)and(5) 
'  for  J/c  and  Mn 

If  either  P,  W,  ls,  or  ^  is  the  unknown  quantity  sought,  the 
criterion  of  (6)  cannot  be  applied,  and  we  .•.  use  both  equa- 
tions in  (6)  and  then  discriminate  between  the  two  results. 
The  greatest  shear  is  Jm^Hlt  in  Fig.  264,  where  12  is 
>  Zi. 

261.  Two  Equal  Terminal  Loads,  Two  Symmetrical  Supports 
Fig.  265.  [Same  case  as  in  Fig.  231,  §  238].  Neglect 
weight  of  beam.  The  reaction  at  each  support  being  =P, 
(from  symmetry),  we  have  for  a  free  body  Om  with  x  <  ^ 

Px      .        .        .     (I) 


while  where  x  >  ^  and  <  Ii 

Px-P  (x—l,}—  ^Lo  .-.  Jf=PZ,        .         (2) 

That  is,  see  (1),  M  varies  directly  with  x  between  0  and  (7, 
while  between  C  and  D  it  is  constant.  Hence  for  safe 
loading 


FLEXURE.      SPECIAL   PROBLEMS. 


305 


The  construction  of  the 
B  moment  diagram  is  evident 
p  from  equations  (1)  and  (2). 

As  for  J,  the  shear,  the 
same  free  bodies  give,  from 
2  (vert,  forces )=0. 

On  OG    .  J=P    ...     (4) 
On  CD    .  J=P—P=zero(5) 


(4)  and  (5)  might  also  be  ob- 
FIG.  265  tained   from  (1)    and  (2)  by 

writing  J=d  M-^-dx,  but  the 

former  method  is  to  be  preferred  in  most  cases,  since  the 
latter  requires  M  to  be  expressed  as  a  function  of  x  while 
the  former  is  applicable  for  examining  separate  sections 
without  making  use  of  a  variable. 

If  the  beam  is  an  I-beam,  the  fact  that  «7is  zero  any- 
where on  G  D  would  indicate  that  we  may  dispense  with 
a  web  along  C  D  to  unite  the  two  flanges ;  but  the  lower 
flange  being  in  compression  and  forming  a  "  long  column  " 
would  tend  to  buckle  out  of  a  straight  line  if  not  stayed  by 
a  web  connection  with  the  other,  or  some  equivalent  brac- 
ing. 

262.  Uniform  Load  over  Part  of  the  Span.  Two  End  Supports. 
Fig.  266.  Let  the  load=  W,  extending  from  one  support 
over  a  portion  =  c,  of  the  span,  (on  the  left,  say,)  so  that 
W=>  we,  w  being  the  load  per  unit  of  length.  Neglect 
weight  of  beam.  For  a  free  body  Om  of  any  length 
x  <  0  B  (i.e.  <  c),  I  momsm=0  gives 


wx* 


2 


(1) 


which  holds  good  for  any  section  on  0  B.  As  for  sections 
on  B  C  it  is  more  simple  to  deal  with  the  free  body  m'C. 
of  length 

x'  <  C  B  from  which  we  have  M=Rt  x'    .     .   (2) 


306 


MECHANICS   OF  ENGINEERING. 


which  shows  the  moment 
curve  for  B  C  to  be  a  straight 
line  DC,  tangent  at  D  to  the 
-T  parabola  (7  D  representing 
eq.  (1.)  (If  there  were  a  con- 
centrated load  at  B,  C  D 
would  meet  the  tangent  at 
D  at  an  angle  instead  of  co- 
inciding with  it  ;  let  the  stu- 
dent show  why,  from  the 
shear  diagram). 

The  shear  for  any  value  of 
x  on  0  B  is  : 


On  OB 

while  on  B  G 


E1  —  wx      .. 
^2=  constant 


(3) 
(4) 


The  shear  diagram  is  constructed  accordingly. 


To  find  the  position  of  the  max.  ordinate  of  the  para- 
bola, (and  this  from  previous  statements  concerning  the 
tangent  at  the  point  D  must  occur  on  0  B,  as  will  be  seen 
and  will  .-.  be  the  Mm  for  the  whole  beam)  we  put  J=0  in 
eq  (3)  whence 


(5) 


and  is  less  than  c,  as  expected.  [The  value  of  Bls=~(l^)t 

=(wc  -7-Z)  (I — £•),  (the  whole  beam  free)  has  been  substi- 
tuted].    This  value  of  x  substituted  in  eq.  (1)  gives 

Jk=(i— # j?. y2 .  we.-.  ^I=y2\\—y2'^-YWc....(Q} 

&  L 

is  the  equation  for  safe  loading. 

The  max.  shear  Jm  is  found  at   0  and  is  =  Ru  which  is 
evidently  >/22,  at  C. 


FLEXURE.      SPECIAL   PROBLEMS. 


307 


263.  Uniform  Load  Over  Whole  Length  With  Two  Symmetrica 
Supports.  Fig.  267. — With  the  notation  expressed  in  the  fig- 
ure, the  following  results  may  be  obtained,  after  having 
divided  the  length  of  the  beam  into  three  parts  for  sepa- 
rate treatment  as  necessitated  by  the  external  forces,  which 
are  the  distributed  load  W,  and 
and  the  two  reactions,  each  = 
Y?,  W.  The  moment  curve  is 
made  up  of  parts  of  three  dis- 
tinct parabolas,  each  with  its 
axis  vertical.  The  central  par- 
abola maj  sink  below  the  hori- 
zontal axis  of  reference  if  the 
supports  are  far  enough  apart, 
in  which  case  (see  Fig.)  the  elas- 
tic curve  of  the  beam  itself  becomes  concave  upward  be- 
tween the  points  E  and  F  of  "  contrary  flexure."  At  each 
of  these  points  the  moment  must  be  zero,  since  the  radius 
of  curvature  is  GO  and  M  =  El  -f-  p  (see  §  231)  at  any  sec- 
tion ;  that  is,  at  these  points  the  moment  curve  crosses  its 
horizontal  axis. 

As  to  the  location  and  amount  of  the  max.  moment  ?Jm, 
inspecting  the  diagram  we  see  that  it  will  be  either  at  Ht 
the  middle,  or  at  both  of  the  supports  B  and  C  (which  from 
symmetry  have  equal  moments),  i.e.,  (with  I  =  total  length,) 


,      R'f]      ( 
and,.-J=j 


either  ^\y4l*-l*]  .....  at  H 


JZ 


at  S  and  G 


according  to  which  is  the  greater  in  any  given  case ;  i.e. 
according  as  ^  is  >  or  <  ^  ^/g. 

The  shear  close  on  the  left  of  S  =  wll}  while  close  to  the 
right  of  B  it  =  #  W  -  wl,.  (It  will  be  noticed  that  in  this 
case  since  the  beam  overhangs,  beyond  the  support,  the 
shear  near  the  support  is  not  equal  to  the  reaction  there, 
as  it  was  in  some  preceding  cases.) 


308 


MECHANICS   OF   ENGINEERING. 


Hence  Jm= 


j  ^  ^_^  J 


according  as 


264,  Hydrostatic  Pressure  Against  a  Vertical  Plank.  —  From 
elementary  hydrostatics  we  know  that  the  pressure,  per 
unit  area,  of  quiescent  water  against  the  vertical  side  of  a 
tank,  varies  directly  with  the  depth,  x,  below  the  surface, 
and  equals  the  weight  of  a  prism  of  water  whose  altitude 
=  x,  and  whose  sectional  area  is  unity.  See  Fig.  268. 


Pis.  268. 

Ttie  plank  is  of  rectangular  cross  section,  its  constant 
breadth,  —  b,  being  r~  to  the  paper,  and  receives  no  sup- 
port except  at  its  two  extremities,  0  and  B,  0  being  level 
with  the  water  surface.  The  loading,  or  pressure,  per  unit 
of  length  of  the  beam,  is  here  variable  and,  by  above  defini- 
nition,  is  =  w=  ?xb,  where  f  =  weight  of  a  cubic  unit 
(i.e.  the  heaviness,  see  §  7)  of  water,  and  x  =  Om  —  depth 
of  any  section  m  below  the  surface.  The  hydrostatic  pres- 
sure on  dx  =  ivdx.  These  pressures  for  equal  dx's,  vary 
as  the  ordinates  of  a  triangle  OR^B. 

Consider  Om  free.  Besides  the  elastic  forces  of  the  ex- 
posed section  m,  the  forces  acting  are  the  reaction  RQ,  and 
the  triangle  of  pressure  OEm.  The  total  of  the  latter  is 


/*. 
wdx  —  fbf  xdx  =' 
'  Jo 


(1) 


and  the  sum  of  the  moments  of  these  pressures  about  m  is 
equal  to  that  of  their  resultant  ( =  their  sum,  since  they 

are  parallel)  about  m,  and  .*. 


fb  — .  .  - 

2      o 


FLEXURE,   SPECIAL  PROBLEMS.          309 

[From  (1)  when  x  =  Z,  we  have  for  the  total  water  pres- 

Z2 
sure  on  the  beam  W{  =  fb  ^  and  since  one-third  of  this 

z 

will  be  borne  at  0  we  have  R0  =^  f&Z2.] 

Now  putting  I(  moms,  about  the  neutral  axis  of  w)=0, 
for  Om  free,  we  have 


(2) 


(which  holds  good  from  x  =  0  to  x  =  1).  From  I  (horiz. 
forces)  =  0  we  have  also  the  shear 

J=S  —  WK=y6  rb  (Z2—  3s2)  ....  (3) 

as  might  also  have  been  obtained  by  differentiating  (2), 
since  J  =  dM  -f-  dx.  By  putting  J  =  0  (§  240,  corollary) 
we  have  for  a  max.  M,  x  =  I  -*•  -^3,  which  is  less  than  I 
and  hence  is  applicable  to  the  problem.  Substitute  this 
in  eq.  2,  and  reduce,  and  we  have 

R'l  R'l    I      1 

v=J*in,i.e.  —  =g-7r-y»    ....  (4) 

as  the  equation  for  safe  loading. 

265.  Example.  —  If  the  thickness  of  the  plank  is  h,  re- 
quired h  =  ?,  if  R'  is  taken  =  1,000  Ibs.  per  sq.  in.  for 
timber  (§  251),  and  I  =  6  feet.  For  the  inch-pound-second 
system  of  units,  we  must  substitute  R'  =  1,000  ;  I  =  72 
inches  ;  r  =  0.036  Ibs.  per  cubic  inch  (heaviness  of  water 
in  this  system  of  units);  while  I  =bh?  -*•  12,  (§  247),  and  e 
=  y2  h.  Hence  from  (4)  we  have 

1000  W    0.0366  x723  ,  _  Q  Q7  . 

12^>P=  -97T"  '  •'•  ~ 

It  will  be  noticed  that  since  x  for  JIfm  =  Z  -5-  Vs,  and  not 
^  Z,  Mm  does  not  occur  in  the  section  opposite  the  resul- 
tant of  the  water  pressure  ;  see  Fig.  268.  The  shear  curve 
is  a  parabola  here  ;  eq.  (3). 


310  MECHANICS   OF   ENGINEERING. 

266.  The  Four  x-Derivatives  of  the  Ordinate  of  the  Elastic  Curve 
— If  y  =  func.  (a;)  is  the  equation  of  the  elastic  curve  for 
any  portion  of  a  loaded  beam,  on  which  portion  the  load 
per  unit  of  length  of  the  beam  is  w  =  either  zero,  (Fig. 
234)  or  =  constant,  (Fig.  235),  or  =  a  continuous  func.  (x) 
(as  in  the  last  §),  we  may  prove,  as  fol- 
lows, that  w  =  the  ie-derivative  of  the 
^  shear.  Fig.  269.  Let  N  and  N'  be  two 
— *•  consecutive  cross-sections  of  a  loaded 
- —  beam,  and  let  the  block  between  them, 
•»—  bearing  its  portion,  wdx,  of  a  distributed 
~  load,  be  considered  free.  The  elastic 
forces  consist  of  the  two  stress-couples 
FIG.  269.  (tensions  and  compressions)  and  the  two 

shears,  e/and  J  -j-  dJt  ck7being  the  shear-increment  conse- 
quent upon  x  receiving  its  increment  dx.  By  putting 
^(vert.  components)  =  0  we  have 

J+dJ—wdx—J=Q  .-.  w=  — 
dx 

Q.  E.  D.     But  J  itself  =  dM  -4-  dx,  (§  240)  and 

M  =  [d?y  -r-  dx2]  EL    By  substitution,  then,  we  have  the 

following  relations : 

y=func.(:r)=  ordinate  at  any  point  of  the  elastic  curve  (1) 
_i=  a  =  slope  at  any  point  of  the  elastic  curve  .  .  (2) 

El  -T^-=  M  =  ordinate  (to  scale)  of  the  moment  curve  (3) 
/73W  i  7-      f  the  ordinate  (to  scale)  )          / 

EI =  '          of  the  shear  '  '  ( 


( the  load  per  unit  of  length   } 

EI  —^  =  w  =  -s  of  beam  =  ordinate  (to  scale)  >      .     (5) 
dx*  |  of  a  curve  of  loading.  ) 

If,  then,  the  equation  of  the  elastic  curve  (the  neutral  line 
of  the  beam  itself ;  a  reality,  and  not  artificial  like  the 


FLEXURE.       SPECIAL   PROBLEMS.  311 

other  curves  spoken  of)  is  given  ;  we  may  by  successive 
differentiation,  for  a  prismatic  and  homogeneous  beam  so 
that  both  E  and  /  are  constant,  find  the  other  four  quan- 
tities mentioned. 

As  to  the  converse  process,  (i.e.  having  given  w  as  a 
function  of  x,  to  find  expressions  for  Jt  M  and  y  as  func- 
tions of  x)  this  is  more  difficult,  since  in  taking  the 
x-anti-derivative,  an  unknown  constant  must  be  added  and 
determined.  The  problem  just  treated  in  §  264,  however, 
offers  a  very  simple  case  since  w  is  the  same  function  of 
x,  along  the  whole  beam,  and  there  is  therefore  but  one  elas- 
tic curve  to  be  determined. 

We  .-.  begin,  numbering  backward,  with 


T?T  _         rfrr  W  =        X'  8e6  (*.a\ 

~dtf~  ~r  ^  (  last  §  and  Fig.  268  (      '        •        ' 

[N.  B.  —  This  derivative  (dJ-r-dx)  is  negative  since  dJ  and 
dx  have  contrary  signs.] 

.-.  (shear=)JE  /&=_  ?  b  ~+  Const. 

But  writing  out  this  equation  for  x=Q,  i.e.  for  the  point 
0,  where  the  shear=  J?0,  we  have  HQ=  0  +  Const.  .-.  Const.  = 
R^  and  hence  write 


.    (Shear)     .    (4a) 
Again  taking  the  x-anti-derivative  of  both  sides 

(Moment  =)  E  I  ^V=—  Tb—+X&+(Con8t.=0)   .    (3o) 
ajr  6 

[At  0,  x=0  alsoM,.-.   Const.  =0].     Again, 


At  0,  where   x=0  dy  -z-dx=a<)=t}ie  unknown  slope  of  the 
elastic  line  at  0,  and  hence  C'=EI  a^ 

•   •   •    (2o) 


312  MECHANICS   OF   ENGINEERING. 

Passing  now  to  y  itself,  and  remembering  that  at  0,  both 
y  and  x  are  zero,  so  that  the  constant,  if  added,  would= 
zero,  we  obtain  (inserting  the  value  of  E^  from  last 


.     (la) 

the  equation  of  the  elastic  curve.  This,  however,  contains 
the  unknown  constant  a,  =  the  slope  at  0.  To  determine 
«0  write  out  eq.  (la)  for  the  point  B,  Fig.  268,  where  x  is 
known  to  be  equal  to  I,  and  y  to  be  =  zero,  solve  for  ap, 
and  insert  its  value  both  in  (la)  and  (2a).  To  find  the 
point  of  max.  y  (i.e.,  of  greatest  deflection)  in  the  elastic 
curve,  write  the  slope,  i.e.  dy  -f-  dx,  =  zero  [see  eq.  2o]  and 
solve  for  x  ;  four  values  will  be  obtained,  of  which  the  one 
lying  between  0  and  I  is  obviously  the  one  to  be  taken. 
This  value  of  x  substituted  in  (la)  will  give  the  maximum 
deflection.  The  location  of  this  maximum  deflection  is 


neither   at   the   centre   of   action   of   the   load 


nor  at  the  section  of  max.  moment  (#  =  £ 

The  qualities  of  the  left  hand  members  of  equations  (1) 
to  (5)  should  be  carefully  noted.  E.  g.,  in  the  inch -pound- 
second  system  of  units  we  should  have  : 

1.  y  (a,  linear  quantity)  =  (so  many)  inches. 

2.  dy-r-dx  (an  abstract  number)  =   (so  many)  abstract 
units. 

3.  M  (a  moment)  ==  (so  many)  inch-pounds. 

4.  J  (a  shear,  i.e.,  force)  =  (so  many)  pounds. 

5.  w  (force  per  linear  unit)  =  (so  many)  pounds  per  run- 
ning inch  of  beam's  length. 

As  to  the  quantities  E,  and  7,  individually,  E  is  pounds 
per  sq.  in.,  and  /has  four  linear  dimensions,  i.e.  (so  many) 
bi-quadratic  inches. 


FLEXUKE       SPECIAL   PROBLEMS.  313 

267.  ResiUence  of  Beam  With  End  Supports.—  Fig.  270.  If  a 
mass  whose  -weight  is  G  (G  large  com- 
pared  with  that  of  beam)  be  allowed  to 
fall  freely  through  a  height  =  h  upon 


PG 


t<.          j  j 

3     T"1"          S  tne  centre  of  a  beam  supported  at  its 

$_J      'pm  extremities,  the  pressure  P  felt  by  the 

FIG.  270.  beam   increases   from   zero   at  the   first 

instant  of  contact  up  to  a  maximum  Pm,  as  already  stated 

in   §233a,  in  which  the  equation  was  derived,  dm  being 

small  compared  with  h, 


The  elastic  limit  is  supposed  not  passed.  In  order  that 
the  maximum  normal  stress  in  any  outer  fibre  shall  at  most 
be=./?',  a  safe  value,  (see  table  §251)  we  must  put 

Tf'T          ~p       I 

—  =  -j—   [according  to  eq.  (2)  §241,]  i.e.  in  equation  (a) 
above,  substitute  Pm=  [4  R'l]  -t-le,  which  gives 

'  '  R"         WV  /M 

'  ~E'^V 

having  put  I=FJc?  (k  being  the  radius  of  gyration  §85) 
and  Fl=  V  the  volume  of  the  (prismatic)  beam.  From 
equation  (b)  we  have  the  energy,  Gh,  (in  ft.-lbs.,  or  inch- 
Ibs.)  of  the  vertical  blow  at  the  middle  which  the  beam  of 
Fig.  270  will  safely  bear,  and  any  one  unknown  quantity 
can  be  computed  from  it,  (but  the  mass  of  G  should  not 
be  small  compared  with  that  of  the  beam.) 

The  energy  of  this  safe  impact,  for  two  beams  of  the 
same  material  and  similar  cross-sections  (similarly  placed), 
is  seen  to  be  proportional  to  their  volumes;  while  if  further- 
more their  cross-sections  are  the  same  and  similarly 
placed,  the  safe  Gh  is  proportional  to  their  lengths.  (These 
same  relations  hold  good,  approximately,  beyond  the  elas 
tic  limit.) 

It  will  be  noticed  that  the  last  statement  is  just  the  con- 


314  MECHANICS   OF   ENGINEERING. 

verse  of  what  was  found  in  §245  for  static  loads,  (the 
pressure  at  the  centre  of  the  beam  being  then  equal  to 
the  weight  of  the  safe  load) ;  for  there  the  longer  the  beam 
(and  /.  the  span)  the  less  the  safe  load,  in  inverse  ratio. 
As  appropriate  in  this  connection,  a  quotation  will  be 
given  from  p.  186  of  "  The  Strength  of  Materials  and 
Structures,"  by  Sir  John  Anderson,  London,  1884,  viz.: 

"  It  appears  from  the  published  experiments  and  state- 
ments of  the  Railway  Commissioners,  that  a  beam  12  feet 
long  will  only  support  ^  of  the  load  that  a  beam  6  feet 
long  of  the  same  breadth  and  depth  will  support,  but  that 
it  will  bear  double  the  weight  suddenly  applied,  as  in  the 
case  of  a  weight  falling  upon  it,"  (from  the  same  height, 
should  be  added) ;  "  or  if  the  same  weights  are  used,  the 
longer  beam  will  not  break  by  the  weight  falling  upon  it 
unless  it  falls  through  twice  the  distance  required  to  frac- 
ture the  shorter  beam." 

268.  Combined  Flexure  and  Torsion.  Crank  Shafts.  Fig.  271. 
Let  OiB  be  the  crank,  and  N0l  the  portion  projecting 

beyond  the  nearest  bearing 
N.  P  is  the  pressure  of  the 
connecting-rod  against  the 
crank-pin  at  a  definite  in- 
stant, the  rotary  motion  be- 
ing uniform.  Let  a=  the 
perpendicular  dropped  from 
the  axis  00l  of  the  shaft 
upon  P,  and  1=  the  distance 
of  P,  along  the  axis  00!  from 
the  cross-section  NmN'  of  the 
shaft,  close  to  the  bearing.  Let  NN'  be  a  diameter  of  this 
section,  and  parallel  to  a.  In  considering  the  portion 
NO^  free,  and  thus  exposing  the  circular  section  NmN', 
we  may  assume  that  the  stresses  to  be  put  in  on  the  ele- 
ments of  this  surface  are  the  tensions  (above  NN'}  and 
the  compressions  (below  NN'}  and  shears  "|  to  NN',  due 
to  the  bending  action  of  P ;  and  the  shearing  stress  tan^ 


FLEXURE.   SPECIAL  PROBLEMS. 


315 


gent  to  the  circles  which  have  0  as  a  common  centre,  and 
pass  through  the  respective  dF's  or  elementary  areas, 
these  latter  stresses  being  due  to  the  twisting  action  of  P. 
In  the  former  set  of  elastic  forces  let  p  =  the  tensile 
stress  per  unit  of  area  in  the  small  parallelopipedical  ele- 
ment m  of  the  helix  which  is  furthest  from  NN'  (the  neu- 
tral axis)  and  /=  the  moment  of  inertia  of  the  circle  about 
NN';  then  taking  moments  about  NN'  for  the  free  body, 
(disregarding  the  motion)  we  have  as  in  cases  of  flexure 
(see  §239) 


[None  of  the  shears  has  a  moment  about  NN'.~\  Next 
taking  moments  about  00X,  (the  flexure  elastic  forces,  both 
normal  and  shearing,  having  no  moments  about  OOj)  we 
have  as  in  torsion  (§216) 


in  which  pf  is  the  shearing  stress  per  unit  of  area,  in  the 
torsional  elastic  forces,  on  any  outermost  dF,  as  at  m; 
and  /p  the  polar  moment  of  inertia  of  the  circle  about  its 
centre  0. 

Next  consider  free,  in  Fig.  272,  a  small  parallelepiped 
taken  from  the  helix  at  m  (of  Fig.  271.)  The  stresses  [see 
§209]  acting  on  the  four  faces  f"  to  the  paper  in  Fig.  272 
are  there  represented,  the  dimensions  (infinitesimal)  being 
n  I  to  NN*,  b  ||  to  00,,  and  d  ~\  to  the  paper  in  Fig.  272. 


./  ^ 


Fig.  278. 


316  MECHANICS   OF   ENGINEERING. 

By  altering  the  ratio  of  b  to  n  we  may  make  the  angle  0 
what  we  please.  It  is  now  proposed  to  consider  free  the 
triangular  prism,  GHT,  to  find  the  intensity  of  normal 
stress  q,  per  unit  of  area,  on  the  diagonal  plane  GH,  (of 
length  =c,)  which  is  a  bounding  face  of  that  triangular 
prism.  See  Fig,  273.  By  writing  2'  (compons.  in  direc- 
tion of  normal  to  GH)=0,  we  shall  have,  transposing, 

qcd=pnd  sin  6+pabd  sin  6-\-p&nd  cos  6  ;    and  solving  for  q 
q=p  ~   sin  d+p&  [|  sin  0+^  .  cos  0]  ;       .      (1) 

but  n  :  c=  sin  6  and  b  :  c=  cos  6    .'. 

q=p  sin20+ps2  sin  0  cos  0        «        .         (2) 

This  may  be  written  (see  eqs.  63  and  60,  O.  W.  J.  Trigo- 
nometry) 

n20        .        .        (3) 


As  the  diagonal  plane  GH  is  taken  in  different  positions 
(i.e.,  as  6  varies),  this  tensile  stress  q  (Ibs.  per  sq.  in.  for 
instance)  also  varies,  being  a  function  of  6,  and  its  max. 
value  may  be  >^>.  To  find  6  for  q  max.  we  put 

-^  ,  =p  sin  20+2pa  cos  20,          .        .      (4) 
=  0,  and  obtain  :    tan  [2(0  for  q  max)]  ---  ^  .    .        .     (5) 

Call  this  value  of  0,  0'.     Since  tan  20'  is  negative,  20'  lies 
either  in  the  second  or  fourth  quadrant,  and  hence 

sin  20'=  ±    ,  ^          and  cos  20'  =  T      .  P  (6) 

vy+4$  f  vy+4pj 

[See  equations  28  and  29  Trigonometry,  O.  W.  J.]     The 


FLEXURE.      CRANK   SHAFT.  337 

upper  signs  refer  to  the  second  quadrant,  the  lower  to  the 
fourth.     If  we  now  differentiate  (4),  obtaining 


.     (7) 


we  note  that  if  the  sine  and  cosine  of  the  [20']  of  the  2nd 
quadrant  [upper  signs  in  (6)]  are  substituted  in  (7)  the  re- 
sult is  negative,  indicating  a  maximum  ;  that  is,  q  is  a  max- 
imum for  0=  the  6'  of  eq.  (6)  when  the  upper  signs  are  taken 
(2nd  quadrant).  To  find  q  max.,  then,  put  ff  for  6  in  (3) 
substituting  from  (6)  (upper  signs).  "We  thus  find 

q  max  =  X[>+Vp2+4p82.]         .         .        (8) 

A  similar  process,  taking  components  parallel  to  GH, 
Fig.  273,  will  yield  qs  max.,  i.e.,  the  max.  shear  per  unit  of 
area,  which  for  a  given  p  and  ps  exists  on  the  diagonal 
plane  GH  in  any  of  its  possible  positions,  as  6  varies. 
This  max.  shearing  stress  is 


»     v.          (9) 

In  the  element  diametrically  opposite  to  ra  in  Fig.  271,  p 
is  compression  instead  of  tension  ;  q  maximum  will  also 
be  compression  but  is  numerically  the  same  as  the  q  max. 
of  eq.  8. 

269.  Example.  —  In  Fig.  271  suppose  P=2  tons  =  4,000 
Ibs.,  a=6  in.,  1=5  in.,  and  that  the  shaft  is  of  wrought 
iron.  Required  its  radius  that  the  max.  tension  or  com- 
pression may  not  exceed  7?'  =12,000  Ibs.  per  sq.  in.;  nor  the 
max.  shear  exceed  S'=  7,000  Ibs.  per  sq.  in.  That  is,  we 
put  y=12,000  in  eq.  (8)  and  solve  for  r  :  also  ys=7,000  in 
(9)  and  solve  for  r.  The  greater  value  of  r  should  be 
taken.  From  equations  (a)  and  (6)  we  have  (see  §§  219  and 
247  for  7P  and  2) 


318  MECHANICS   OF   ENGINEERING. 


which  in  (8)  and  (9)  give 

max.  q=y2*L  [4Z+4/(4^)a+4(2a)a]     .     .      (8a) 

KIT 

n 

and  max.  qt=%— s\/(4Z)2+4(2a)2    •        •        •  W 


With  max.  g=12,000,  and  the  values  of  P,  a,  and  Z,  already 
given,  (units,  inch  and  pound)  we  have  from  (8a),  ^=2.72 
cubic  inches  .-.  r=1.39  inches. 

Next,  with  max.  gs=7,000 ;  P,  a,  and  I  as  before ;  from 
(9a),  r*=%.84:  cubic  inches  .*.  r=1.41  inches. 

The  latter  value  of  r,  1.41  inches,  should  be  adopted.  It 
is  here  supposed  that  the  crank-pin  is  in  such  a  position 
(when  P=4,000  Ibs.,  and  a=6  in.)  that  q  max.  (and  qs 
max.)  are  greater  than  for  any  other  position  ;  a  number 
of  trials  may  be  necessary  to  decide  this,  since  P  and  a  are 
different  with  each  new  position  of  the  connecting  rod.  If 
the  shaft  and  its  connections  are  exposed  to  shocks,  R  and 
JS'  should  be  taken  much  smaller. 

270.  Another  Example  of  combined  torsion  and  flexure  is 

shown  in  Fig.   274.      The 

P°1     'r/7^\   ^~~"^k~~          p»    work  °^  *ke  working  force 
oj-^//   i_J\     /T\~1''     1    pi(vertical  cog-pressure)  is 
y^\      tv^T^Tri^ij  <t^~— — -JU  expended  in  overcoming  the 

resistance  (another  vertical 
cog-pressure)  Qlt 

>-2?4.  That  is,  the  rigid  body 

consisting  of  the  two  wheels  and  shaft  is  employed  to 
transmit  power,  at  a  uniform  angular  velocity,  and  since 
it  is  symmetrical  about  its  axis  of  rotation  the  forces  act- 
ing on  it,  considered  free,  form  a  balanced  system.  (See 
§  114).  Hence  given  PI  and  the  various  geometrical  quan- 


FLEXURE.       CRANK   SHAFT.  319 

titles  «!,  &i,  etc.,  we  may  obtain  $„  and  the  reactions  P0  and 
PB,  in  terms  of  Pt.  The  greatest  moment  of  flexure  in  the 
shaft  will  be  either  P0Z,,  at  (7;  or  PB^,  at  D.  The  portion 
CD  is  under  torsion,  of  a  moment  of  torsion  =P1a1=  Qfa. 
Hence  we  proceed  as  in  the  example  of  §  269,  simply  put- 
ting P0li  (or  PB?3,  whichever  is  the  greater)  in  place  of  PZ, 
and  P&i  in  place  of  Pa.  We  have  here  neglected  the 
weight  of  the  shaft  and  wheels.  If  Ql  were  an  upward  ver- 
tical force  and  hence  on  the  same  side  of  the  shaft  as  Plt 
the  reactions  P0  and  PB  would  be  less  than  before,  and  one 
or  both  of  them  might  be  reversed  in  direction. 


320  MECHANICS   OF   ENGINEERING. 


CHAPTEK    IV. 


FLEXURE,    CONTINUED, 


CONTINUOUS  GIRDERS. 


271.  Definition. — A  continuous  girder,  for  present  pur- 
poses, may  be  denned  to  be  a  loaded  straight  beam  sup- 
ported in  more  than  two  points,  in  which  case  we  can  no 
longer,  as  heretofore,  determine  the  reactions  at  the  sup- 
ports from  simple  Statics  alone,  but  must  have  recourse 
io  the  equations  of  the  several  elastic  curves  formed  by  its 
neutral  line,  which  equations  involve  directly  or  indirect- 
ly the  reactions  sought ;  the  latter  may  then  be  found  as 
if  they  were  constants  of  integration.  Practically  this 
amounts  to  saying  that  the  reactions  depend  on  the  man- 
ner in  which  the  beam  bends  ;  whereas  in  previous  cases, 
with  only  two  supports,  the  reactions  were  independent  of 
the  forms  of  the  elastic  curves  (the  flexure  being  slight, 
however). 

As  an  Illustration,  if  the  straight  beam  of  Fig.  275  is  placed 
on  three  supports  6>,  B,  and  (?,  at  the  same  level,  the 
reactions  of  these  supports  seem  at  first  sight  indeterm- 
inate ;  for  on  considering  the  p ./ H 

whole  beam  free,  we  have  three     rH3^^H~~^     »         ^ 

unknown  quantities  and  only  B/^ — f-~     A0^"""""f~""'''''2A 
two   equations,  viz :    S    (vert.  '    Fl0  275 

compons.)  —  0  and  S  (moms,  about  some  point")  =  0.     If 
now  0  be  gradually  lowered,  it  receives  less  and  less  pres- 


FLEXURE.       CONTIGUOUS   GIRDERS.  321 

sure,  until  it  finally  reaches  a  position  where  the  beam 
barely  touches  it ;  and  then  O's  reaction  is  zero,  and  B  and 
C  support  the  beam  as  if  0  were  not  there.  As  to  how 
low  0  must  sink  to  obtain  this  position,  depends  on  the 
,  stiffness  and  load  of  the  beam.  Again,  if  0  be  raised 
above  the  level  of  B  and  C  it  receives  greater  and  greater 
pressure,  until  the  beam  fails  to  touch  one  of  the  other 
supports.  Still  another  consideration  is  that  if  the  beam 
were  tapering  in  form,  being  stiffest  at  0,  and  pointed  at 
B  and  6y,  the  three  reactions  would  be  different  from  their 
values  for  a  prismatic  beam.  It  is  therefore  evident  that 
for  more  than  two  supports  the  values  of  the  reactions  de- 
pend on  the  relative  heights  of  the  supports  and  upon  the 
form  and  elasticity  of  the  beam,  as  well  as  upon  the  load. 
The  circumstance  that  the  beam  is  made  continuous  over 
the  support  0,  instead  of  being  cut  apart  at  0  into  two 
independent  beams,  each  covering  its  own  span  and  hav- 
ing its  own  two  supports,  shows  the  significance  of  the 
term  "  continuous  girder." 

All  the  cases  here  considered  will  be  comparatively 
simple,  from  the  symmetry  of  their  conditions.  The 
beams  will  all  be  prismatic,  and  all  external  forces  (i.e. 
loads  and  reactions)  perpendicular  to  the  beam  and  in  the 
same  plane.  All  supports  at  the  same  level 

272.  Two  Equal  Spans ;  Two  Concentrated  Loads,  One  in  the  Mid- 
dle of  Each  Span.  Prismatic  Beam. — Fig.  275.  Let  each  half- 
span  =  y2  I.  Neglect  the  weight  of  the  beam.  Required 
the  reactions  of  the  three  supports.  Call  them  PB,  P0  and 
Pc.  From  symmetry  PB  =  Pc,  and  the  tangent  to  the 
elastic  curve  at  0  is  horizontal ;  and  since  the  supports 
are  on  a  level  the  deflection  of  C  (and  B}  below  O's  tangent 
is  zero.  The  separate  elastic  curves  OD  and  DC  have  a 
common  slope  and  a  common  ordinate  at  D. 

For  the  equation  of  OD,  make  a  section  n  anywhere  be- 
tween 0  and  D,  considering  nC  a  free  body.  Fig.  276  (a) 


322 


MECHANICS  OF  ENGINEERING. 


PC  |Y 


Fie.  876. 


with  origin  and  axis  as  there  indicated.      From  2  (moms 
about  neutral  axis  of  ri)  =  0  we  have  (see  §  231) 


(1 


=V)    .    (2) 


The  constant  =  0,  for  at  0  both  x,  and  dy  -*-  dx,  =  0. 
Taking  the  x-anti-derivative  of  (2)  we  have 


(B) 


Here  again  the  constant  is  zero  since  at  O,x  and  y  both  =0. 
(3)  is  the  equation  of  OD,  and  allows  no  value  of  x  <0 
or>  -A.  It  contains  the  unknown  force  Pc. 

For  the  equation  of  DC,  let  the  variable  section  n  be  made 
anywhere  between  D  and  C,  and  we  have  (Fig.  276  (V)  ;  x 
may  now  range  between  y2l  and  I) 


(5)' 


To  determine  C",  put  x  =  y2l  both  in  (5)'  and  (2),  and 
equate  the  results  (for  the  two  curves  have  a  common 
tangent  line  at  D)  whence  C'  —  %  PI? 


(5) 


FLEXURE.      CONTINUOUS   GIRDERS.  31*3 

Hence    Ely    =  #  PPx-Pc       !_      +  (7"        .        .      (6)' 


At  D  the  curves  have  the  same  y,  hence  put  x  =  -L  in  the 
right  hand  member  both  of  (3)  and  (6)',  equating  results, 
and  we  derive  C"'=  —  1  PP 


(6) 


which  is  the  equation  of  DC,  but  contains  the  unknown 
reaction  Pc.  To  determine  Pc  we  employ  the  fact  that  O's 
tangent  passes  through  C,  (supports  on  same  level)  and 
hence  when  x  =  I  in  (6),  y  is  known  to  be  zero.  Making 
these  substitutions  in  (6)  we  have 


From  symmetry  PB  also  =  -JP,  while  P0  must  =  ~P, 
since  PB  +  P0  +  Pc  =  2  P  (whole  beam  free).  [NoTE.  — 
If  the  supports  were  not  on  a  level,  but  if,  (for  instance) 
the  middle  support  0  were  a  small  distance  =  ^0  below 
the  level  line  joining  the  others,  we  should  put  x  =  I  and 
y  =  —  ho  in  eq.  (6),  and  thus  obtain  PB  =  Pc=  -?6  P  + 

3EI^,  which  depends  on  the  material   and  form   of  the 

C 

prismatic  beam  and  upon  the  length  of  one  span,  (whereas 
with  supports  all  on  a  levd,  PB  =  Pc  =  A  P  is  independent 
of  the  material  and  form  of  the  beam  so  long  as  it  is  ho- 
mogeneous and  prismatic.)  If  P0,  which  would  then  = 
g  P  —  6  J577(Ao-hP),  is  found  to  be  negative,  it  shows  that 
0  requires  a  support  from  above,  instead  of  below,  to 
cause  it  to  occupy  a  position  Jio  below  the  other  supports, 
i.e.  the  beam  must  be  "  latched  down  "  at  0.] 

The  moment  diagram  of  this  case  can  now  be  easily  con- 
structed ;  Fig.  277.  For  any  free  body  nG,  n  lying  in  DC, 
we  have 


324 


MECHANICS   OF  ENGINEERING. 


i.e.,  varies  directly  as  x,  un- 
c  til  x  passes  D  when,  for  any 
point  on  DO, 

M=*/l6Px-P(x-1-) 

which  is  =0,  (point  of  in- 
flection   of    elastic    curve) 
for  x=8/n  I  (note  that  x  is 
PIS.  277.  measured  from    C  in  this 

figure)  and  at  0,  where x=l, becomes  — |,-PZ 

Hence,  since  M  max.  =^Pl,  the  equation  for  safe  loading 

is 

6 


^      p/ 

~e~     32 


•    (7) 


The  shear  at  C  and  anywhere  on  CD=^P,  while  on  DO 
it  =l^P  in  the  opposite  direction 

•:<rm-$P        ....     (8) 

The  moment  and  shear  diagrams  are  easily  constructed, 
as  shown  in  Fig.  277,  the  former  being  svmmetrical  about 
a  vertical  line  through  0,  the  latter  about  the  point  0" 
Both  are  bounded  by  right  lines. 

273.    Two  Equal   Spans.    Uniformly   Distributed   Load   Over 

Whole  Length.  Prismatic  Beam, 
c   —Fig.  278.    Supports  B,  0, 
C,   on  a  level.     Total  load 
^  =  2  W=  %wl  and  may  include 
»(*:»)      I  that  of  the  beam  ;  w  is  con- 
\\\\  |  j  stant.  As  before,  from  sym- 
metry PB=  Pc,  the  unknown 
reactions   at    the   extremi- 
3. 278.  ties. 


FLEXURE.      CONTINUOUS  GIRDERS.  325 

Let  On=x  ;  then  with  n C  free,  2  moms,  about  w=0  gives 


=0]  (2) 
[Const.  =0  for  at  0  both  dy-t-dx  the  slope,  and  x,  are  =0] 

=ty  (3) 


[Const.  =0  for  at  0  both  x  andy  are  =0].  Equations  (1), 
(2),  and  (3)  admit  of  any  value  of  x  from  0  to  I,  i.e.,  hold 
good  for  any  point  of  the  elastic  curve  0(7,  the  loading  on 
which  follows  a  continuous  law  (viz.  :  w=  constant).  But 
when  x=l,  i.e.,  at  (7,  y  is  known  to  be  equal  to  zero,  since 
0,  B  and  (7  are  on  the  axis  of  X,  (tangent  at  0).  With 
these  values  of  x  and  y  in  eq.  (3)  we  have 


The  Moment  and  Shear  Diagrams  can  now  be  formed  since 
M  all  the    external    forces   are 


AW  known. 
I  I        ure  x 


In 


Fig.   279  meas- 
.     Then  in  any 
section  n  the  moment  of  the 
I  "  stress-couple  "  is 


.  (1) 


1  which  holds  good  for  any 
value  of  x  on  CO,  i.e.,  from 
a;=0  up  to  x=l.  By  inspec- 
tion it  is  seen  that  for  x=0, 
M=0  ;  and  also  for  x=ffl,  M=0,  at  the  inflection  point  G, 
beyond  which,  toward  0,  the  upper  fibres  are  in  tension 


326  MECHANICS   OF   ENGINEERING. 

the  lower  in  compression,  whereas  between  C  and  G  they 
are  vice  versa.  As  to  the  greatest  moment  to  be  found  on 
CG,  put  dM-^-dx—Q  and  solve  for  x.  This  gives 

y%  W—wx=0  .-.  [x  for  M  max.]  =  y&        .       (2) 
which  in  eq.  (1)  gives 

±Wl        .         .      (2) 


But  this  is  numerically  less  than  M0(=  —  ^  WT)  hence  the 
stress  in  the  outer  fibre  at  0  being 

P0=X^,     ••        •       •        (3) 

the  equation  for  safe  loading.  is 

.        .        .        .    (4) 


the  same  as  if  the  beam  were  cut  through  at  0,  each  half, 
of  length  I,  retaining  the  same  load  as  before  [see  §  242  eq. 
(2)].  Hence  making  the  girder  continuous  over  the  mid- 
dle support  does  not  make  it  any  stronger  under  a  uni- 
formly distributed  load  ;  but  it  does  make  it  considerably 
stiver. 

As  for  the  shear,  «7,  we  obtain  it  for  any  section  by  tak- 
ing the  x-derivative  of  M  in  eq.  (1),  or  by  putting  ^'(ver- 
tical forces)  =0  for  the  free  body  nC,  and  thus  have  for 
any  section  on  CO 

J=3/8W—  wx     .  ,      ..        .          (5) 

J  is  zero  for  x  =  ^l  (where  M  reaches  its  calculus  maxi- 
mum J^N  ;  see  above)  and  for  x=l  it  =—%  W  which  is  nu 
merically  greater  than  ^  W,  its  value  at  C.  Hence 

....      (6) 


FLEXURE.      CONTINUOUS   GIRDERS.  -'527 

The  moment  curve  is  a  parabola  (a  separate  one  for  each 
span),  the  shear  curve  a  straight  line,  inclined  to  the  hor- 
izontal, for  each  span. 

Problem. — How  would  the  reactions  in  Fig.  278  be 
changed  if  the  support  0  were  lowered  a  (small)  distance 
ht)  below  the  level  of  the  other  two  ? 


274.    Prismatic  Beam  Fixed   Horizontally  at  Both  Ends  (at 
Same  Level).    Single  Load  at  Middle.— Fig.    280.     [As  usual 

the  beam  is  understood  to 
be  homogeneous  so  that  E 
is  the  same  at  all  sections]. 
The  building  in,  or  fixing, 
of  the  two  ends  is  supposed 
to  be  of  such  a  nature  as  to 
cause  no  horizontal  con- 
Fio.280.  straint ;  i.e.,  the  beam  does 

not  act  as  a  cord  or  chain,  in  any  manner,  and  hence  the 
sum  of  the  horizontal  components  of  the  stresses  in  any 
section  is  zero,  as  in  all  preceding  cases  of  flexure.  In 
other  words  the  neutral  axis  still  contains  the  centre  of 
gravity  of  the  section  and  the  tensions  and  compressions 
are  equivalent  to  a  couple  (the  stress-couple)  whose  mo- 
ment is  the  "  moment  of  flexure." 

If  the  beam  is  conceived  cut  through  close  to  both  wall 
faces,  and  this  portion  of  length=Z,  considered  free,  the 
forces  holding  it  in  equilibrium  consist  of  the  downward 
force  P  (the  load) ;  two  upward  shears  J0  and  Jc  (one  at 
each  section) ;  and  two  "  stress-couples  "  one  in  each  sec- 
tion, whose  moments  are  Mn  and  Mc.  From  symmetry  we 
know  that  J,,=  JC,  and  that  M0=MC.  From  2  F=0  for  the 
free  body  just  mentioned,  (but  not  shown  in  the  figure), 
and  from  symmetry,  we  have  Jn=  ^4  P  and  «/c=  y?.  P  ',  but 
to  determine  M.t  and  Mr,  the  form  of  the  elastic  curves 
0  B  and  B  C  must  be  taken  into  account  as  follows  : 
Equation  of  OB,  Fig.  280.  I  [mom.  about  neutral  axis 
of  any  section  n  on  0  5]  =  0  (for  the  free  body  nC  which 


328 


MECHANICS   OF   ENGINEERING. 


has  a  section  exposed  at  each  end,  n  being  the  variable 
section)  will  give 


-=  P(y2  l-x)  +  M—  %  P  (l—x) 

* 


(1) 


[Note.  In  forming  this  moment  equation,  notice  that 
M,.  is  the  sum  of  the  moments  of  the  tensions  and  com- 
pressions at  C  about  the  neutral  axis  at  n,  just  as  much  as 
about  the  neutral  axis  of  0;  for  those  tensions  and  com- 
pressions are  equivalent  to  a  couple,  and  hence  the  sum  of 
their  moments  is  the  same  taken  about  any  axis  whatever 
"|  to  the  plane  of  the  couple  (§32).] 

Taking  the  x-anti-derivative  of  each  member  of  (1), 


I x— 


-/-' 


P(lx-} 


(2) 


(The  constant  is  not  expressed,  as  it  is  zero).  Now  from 
symmetry  we  know  that  the  tangent-line  to  the  curve  0  B 
at  B  is  horizontal,  i.e.,  for  x=  ^l,  dy-s-dx=Q,  and  these 
values  in  eq.  (2)  give  us 

0=  y8  PP+  y2MJ—*-PV;  whence  M,= M,=%  PI     .     (3) 

Safe  Loading.  Fig.  281.  Having  now  all  the  forces  which 
act  as  external  forces  in  straining  the  beam  OC,  we  are 
ready  to  draw  the  moment  diagram  and  find  Mm.  For  con- 
venience measure  x  from  (7.  For  the  free  body  nC,  we 
have  [see  eq.  (3)] 


r=i^  Pl-y2Px 


(4) 


FIG.  281. 


Eq.  (4)  holds  good  for  any 
section  on  CB.  By  put- 

7"  ting  x=0  we  have  M=MC— 
%  PI;  lay  off  HC'=MC  to 
scale  (so  many  inch-pounds 

c  moment  to  the  inch  of  pa- 
per). At  B,  for  0?=^  ly 
ME= — y%  PI ;  hence  lay 
off  B'D=y&  PI  on  the  op- 
posite side  of  the  axis  O'G' 


FLEXURE.       CONTINUOUS    GIRDERS. 


329 


from  HC',  and  join  DH.  DK,  symmetrical  with  DH  about 
B' D,  completes  the  moment  curves,  viz.:  two  right  lines. 
The  max.  M  is  evidently  =yi  PI  and  the  equation  of  safe 
loading 


=      PI 


(5) 


Hence  the  beam  is  twice  as  strong  as  if  simply  supported 
at  the  ends,  under  this  load  ;  it  may  also  be  proved  to  be 
four  times  as  stiff. 

The  points  of  inflection  of  the  elastic  curve  are  in  the 
middles  of  the  half-spans,  while  the  max.  shear  is 


(6) 


275.  Prismatic  Beam  Fixed  Horizontally  at  Both  Ends  [at  Same 
Level].  Uniformly  Distributed  Load  Over  the  Whole  Length. 
Fig.  282.  As  in  the  preceding  problem,  we  know  from 
symmetry  that  J0=Jc=y2W~y2  ivl,  and  that  M0=MC,  and 
determine  the  latter  quantities  by  the  equation  of  the 
curve  OCy  there  being  but  one  curve  in  the  present  in- 
stance, instead  of  two,  as  there  is  no  change  in  the  law  of 
loading  between  O  and  C.  With  nO  free,  I  (momn)=0 
gives 


and  .*. 


(1) 


(2) 


FIG.  282. 


330 


MECHANICS   OF   ENGINEERING. 


The  tangent  line  at  0  being  horizontal  we  have  for  x=Q,—/  = 

cLx 

0,  /.  (7=0.     But  since  the  tangent  line  at  C  is  also  hori- 
zontal, we  may  for  x=l  put  dy-±dx=(),  and  obtain 


;  whence  1/0=A-  Wl 


(3) 


as  the  moment  of  the  stress-couple  close  to  the  wall  at  0 
and  at  C. 

Hence,  Fig.  283,  the  equation  of  the  moment  curve  (a 
single  continuous  curve  in  this  case)  is  found  by  putting 
2'  (momn)=0  for  the  free  body  nO,  of  length  x,  thus 
obtaining 

1*.. 

\N-wl 

LLUJJ  I  i  LUJi 


FIG. 


= 


(4) 


an  equation  of  the  second  degree,  indicating  a  conic.  At  0, 
M=M0  of  course,=  l.  Wl ;  at.Z?by  putting  x=]/2  I  in  (4),  we 
have  MH= — 2\  ^>  which  is  less  than  M0,  although  MB  is  the 
calculus  max.  (negative)  for  M,  as  may  be  shown  by  writ- 
ing the  expression  for  the  shear  (J=%  W- — wx)  equal  to 
zero,  etc. 


FLEXURE.       CONTINUOUS    GIUDERS.  331 

Hence  Mm=^  Wl,  and  the  equation  for  safe  loading  is 

^=r^  ^1 (5) 

Since  (with  this  form  of  loading)  if  the  beam  were  not 
built  in  but  simply  rested  on  two  end  supports,  the  equa- 
tion for  safe  loading  would  be  [R'I-t-e~]  =  *£  Wl,  (see  §242), 
it  is  evident  that  with  the  present  mode  of  support  it  is  50 
per  cent,  stronger  as  compared  with  the  other  ;  i.e.,  as  re- 
gards normal  stresses  in  the  outer  elements.  As  regards 
shearing  stresses  in  the  web  if  it  has  one,  it  is  no  stronger* 
since  t7m=y£  Win  both  cases. 

As  to  stiffness  under  the  uniform  load,  the  max.  deflec- 
tion in  the  present  case  may  be  shown  to  be  only  -i-  of  that 
in  the  case  of  the  simple  end  supports.  u~~ 

It  is  noteworthy  that  the  shear  diagram  in  Fig.  283  is 
identical  with  that  for  simple  end  supports  §242,  under 
uniform  load  ;  while  the  moment  diagrams  differ  as  fol- 
lows :  The  parabola  KB' A,  Fig.  283,  is  identical  with  that 
in  Fig.  235,  but  the  horizontal  axis  from  which  the  ordi- 
nates  of  the  former  are  measured,  instead  of  joining  the 
extremities  of  the  curve,  cuts  it  in  such  a  way  as  to  have 
equal  areas  between  it  and  the  curve,  on  opposite  sides 

i.e.,  areas  \_KC'H>+AG'0']=vxQ&  H'G'B' 

In  other  words,  the  effect  of  fixing  the  ends  horizontally 
is  to  shift  the  moment  parabola  upward  a  distance  =  M0 
(to  scale),  =  ~  Wl,  with  regard  to  the  axis  of  reference, 
O'B',  in  Fig.  235. 

276.  Remarks. — The  foregoing  very  simple  cases  of  con- 
tinuous girders  illustrate  the  means  employed  for  deter- 
mining the  reactions  of  supports  and  eventually  the  max. 
moment  and  the  equations  for  safe  loading  and  for  deflec- 
tions. When  there  are  more  than  three  supports,  with 
spans  of  unequal  length,  and  loading  of  any  description, 
the  analysis  leading  to  the  above  results  is  much  more 
complicated  and  tedious,  but  is  considerably  simplified 


332  MECHANICS   OF    ENGINEERING. 

and  systematized  by  the  use  of  the  remarkable  theorem  of 
three  moments,  the  discovery  of  Clapeyron,  in  1857.  By 
this  theorem,  given  the  spans,  the  loading,  and  the  vertical 
heights  of  the  supports,  we  are  enabled  to  write  out  a  rela- 
tion between  the  moments  of  each  three  consecutive  sup- 
ports, and  thus  obtain  a  sufficient  number  of  equations  to 
determine  the  moments  at  all  the  supports  [p.  641  Rankine's 
Applied  Mechanics.]  From  these  moments  the  shears 
close  to  each  side  of  each  support  are  found,  then  the 
reactions,  and  from  these  and  the  given  loads  the  moment 
at  any  section  can  be  determined ;  and  hence  finally  the 
max.  moment  Mm,  and  the  max.  shear  Jm. 

The  treatment  of  the  general  case  of  continuous  girders 
by  graphic  methods,  however,  is  comparatively  simple, 
and  its  presentation  is  therefore  deferred,  §  391. 


THE    DANGEROUS    SECTION    OF   NON-PRIS- 
MATIC   BEAMS. 

277.  Remarks.  By  "  dangerous  section  "  is  meant  that  sec- 
tion (in  a  given  beam  under  given  loading  with  given  mode 
of  support)  where  p,  the  normal  stress  in  the  outer  fibre, 
at  distance  e  from  its  neutral  axis,  is  greater  than  in  the 
outer  fibre  of  any  other  section.  Hence  the  elasticity  of 
the  material  will  be  first  impaired  in  the  outer  fibre  of 
this  section,  if  the  load  is  gradually  increased  in  amount 
(but  not  altered  in  distribution). 

In  all  preceding  problems,  the  beam  being  prismatic,  /, 
the  moment  of  inertia,  and  e  were  the  same  in  all  sections, 

hence  when  the  equation    ¥-=M    [§239]  was  solved  for  p, 

€> 

giving 


we  found  that  p  was  a  max.,  =  pm,  for  that  section  whose 
M  was  a  maximum,  since  p  varied  as  M,  for  the  moment 


FLEXURE       XON-PRISMATIC   BEAMS.  333 

of  the  stress-couple,  as  successive  sections  along  the  beam 
were  examined. 

But  for  a  non-prismatic  beam  /and  e  change,  from  sec- 
tion to  section,  as  well  as  M,  and  the  ordinate  of  the 
moment  diagram  no  longer  shows  the  variation  of  p,  nor 
is  p  a  max.  where  M  is  a  max.  To  find  the  dangerous 
section,  then,  for  a  non-prismatic  beam,  we  express  the  J/, 
the  /,  and  the  e  of  any  section  in  terms  of  x,  thus  obtain- 
ing j3=func.  (x),  then  writing  dp-t-dx=Q,  and  solving  for  x. 

278.  Dangerous  Section  in  a  Double  Truncated  Wedge.  Two 
End  Supports.  Single  Load  in  Middle. — The  form  is  shown  in 
Fig.  284.  Neglect  weight  of  beam  ;  measure  x  from  one  sup- 
port 0.  The 
t  reaction  at 
r_~^r4  7~j[w~jDl  each  support 
is  y2  P.  The 
width  of  the 

Fio.284.  beam  -   b  at 

all  sections,  while  its  height,  v,  varies,  being  =  h  at  0. 
To  express  the  e  =  y2  v,  and  the  /  =  1  btf  (§247)  of  any 
section  on  OC,  in  terms  of  x,  conceive  the  sloping  faces 
of  the  truncated  wedge  to  be  prolonged  to  their  intersec- 
tion A,  at  a  known  distance  =  c  from  the  face  at  0.  We 
then  have  from  similar  triangles 

v  :  x   +  c  :  :  h  :  c,  .-.  v  =  ~  (x  +  c)      .     .     (1) 

C 

and  .-.  e  =  y2  -  (x+c)  and  1=  1  6  -O+c]s     .     (2) 
c  12      c' 

For  the  free  body  nO,  I  (moms.n)  =  0  gives 

[That  is,  the  M  =  y2  Px.]     But  from  (2),  (3)  becomes 


By  putting  dp  -*-  dx  ==  0  we  obtain  both  x  =  —  c,  and 


334 


MECHANICS   OF   ENGINEERING. 


x  =  -f  c,  of  which  the  latter,  x  =  +  c,  corresponds  to  a 
maximum  for  p  (since  it  will  be  found  to  give  a  negative 
result  on  substitution  in  d*p  -4-  dx*). 

Hence  the  dangerous  section  is  as  far  from  the  support 
0,  as  the  imaginary  edge,  A,  of  the  completed  wedge,  but 
of  course  on  the  opposite  side.  This  supposes  that  the 
half -span,  )4l,  is  >  c\  if  not,  the  dangerous  section  will 
be  at  the  middle  of  the  beam,  as  if  the  beam  were 
prismatic. 

He-    with   \***^**^*H>^ 

)  at  middle)  ( 

while  with  )  fhe  Aquation  for  safe   (  R^  [2A]2 

/7  V  loading  is  :  (put  x=c  <  i — i-=  y2  PC     (6) 

jand|>=;.B'ia[3])         ( 

(see  §239.) 

279.    Double  Truncated  Pyramid  and  Cone.     Fig.  285.     For 


FIG.  285. 

the  truncated  pyramid  both  width  —  u,  and  height  =  v, 
are  variable,  and  if  b  and  h  are  the  dimensions  at  0,  and 
c  =  OA  —  distance  from  0  to  the  imaginary  vertex  A,  we 

shall  have  from  similar  triangles  u=~  (a?+c)and  v=  -(cc+c). 

in  the  moment 


Hence,  substituting  e=}4v  and  I=L 
equation 


.    .       (7) 


,~ 

'      '     ' 


FLEXURE.      NON-PKISMATIC   BEAMS.  335 

Putting  this  =  0,  we  have  x  =  —  c,  x  =  —  c,  and  x  = 
+  y2  c,  hence  the  dangerous  section  is  at  a  distance  x  =  y2  c 
from  0,  and  the  equation  for  safe  loading  is 

either  *?*™L->  ft  Fl  .  .  .  .  if  #  Z  is  <  */2  c       ....    (9) 
b 

(in  which  b'  and  A'  are  the  dimensions  at  mid-span) 


.  . 

For  the  truncated  cone  (see  Fig.  285  also,  on  right)  where 
e  =  the   variable  radius   r,  and  /  =  %  it  r*,  we  also  have 

/  =[Const.]    .  ^-f    ....     (11) 

and  hence  p  is  a  max.  for  x  =  }4  c,  and  the  equation  for 
safe  loading 


either  =  %  Pl,tor  %l  <  %  c     .....     (12) 

(where  r'  =  radius  of  mid  -span  section)  ; 


or  x  =  %  pCf  for  I/4  i  > 

(where  r0  =  radius  of  extremity.) 


NON-PRISMATIC    BEAMS    OF    "UNIFORM 
STRENGTH." 

280.  Remarks.  A  beam  is  said  to  be  of "  uniform 
strength  "  when  its  form,  its  mode  of  support,  and  the  dis- 
tribution of  loading,  are  such  that  the  normal  stress  p  has 
the  same  value  in  all  the  outer  fibres,  and  thus  one  ele- 
ment of  economy  is  secured,  viz. :  that  all  the  outer  fibres 
may  be  made  to  do  full  duty,  since  under  the  safe  loading, 
p  will  be  =  to  R'  in  all  of  them.  [Of  course,  in  all  cases 
of  flexure,  the  elements  between  the  neutral  surface  and 


336 


MECHANICS   OF   ENGINEERING. 


the  outer  fibres  being  under  tensions  and  compressions 
less  than  R'  per  sq.  inch,  are  not  doing  full  duty,  as 
regards  economy  of  material,  unless  perhaps  with  respect 
to  shearing  stresses.]  In  Fig.  265,  §261,  we  have  already 
had  an  instance  of  a  body  of  uniform  strength  in  flexure, 
viz. :  the  middle  segment,  CD,  of  that  figure ;  for  the 
moment  is  the  same  for  all  sections  of  CD  [eq.  (2)  of  that 
§],  and  hence  the  normal  stress  p  in  the  outer  fibres  (the 
beam  being  prismatic  in  that  instance). 

In  the  following  problems  the  weight  of  the  beam  itself 
is  neglected.  The  general  method  pursued  will  be  to  find 
an  expression  for  the  outer -fibre-stress  p,  at  a  definite  sec- 
tion of  the  beam,  where  the  dimensions  of  the  section  are 
known  or  assumed,  then  an  expression  f or  p  in  the  varia- 
ble section,  and  equate  the  two.  For  clearness  the  figures 
are  exaggerated,  vertically. 


281.    Parabolic  Working  Beam.    UnsymmetricaL      Fig. 


FIG.  886. 

CBO  is  a  working  beam  or  lever,  B  being  the  fixed  fulcrum 
or  bearing.  The  force  P0  being  given  we  may  compute  Pc 
from  the  mom.  equation  P010  —  PJn  while  the  fulcrum 
reaction  is  PB=P0-\-Pc.  All  the  forces  are  ~[  to  the  beam. 
The  beam  is  to  have  the  same  width  b  at  all  points,  and  is 
to  be  rectangular  in  section. 

Eequired  first,  the  proper  height  ^,  at  B,  for  safety. 
From  the  free  body  BO,  of  length  =  10,  we  have  I  (momsB) 
=  0;  i.e., 


(1) 


FLEXURE.       NOX-PRISMATIC   BEAMS.  337 

Hence,  putting  pB  =  R',  ht  becomes  known  from  (1). 

Required,  secondly,  the  relation  between  the  variable 
height  v  (at  any  section  n)  and  the  distance  x  of  n  from  0. 
For  the  free  body  nO,  we  have  (2'  momsa  =  0) 

pJ=PcX .  or  P«±M>  =Pj,  ^a ...  ft 


But  for  "  uniform  strength "  pa  must  =  pa  \  hence 
equate  their  values  from  (1)  and  (2)  and  we  have 

5  =  A,  which  may  be  written  (%  vf  =  MW    x    (3) 

so  as  to  make  the  relation  between  the  abscissa  x  and  the 
ordinate  }4  v  more  marked ;  it  is  the  equation  of  a  para- 
bola, whose  vertex  is  at  0. 

The  parabolic  outline  for  the  portion  BC  is  found  simi- 
larly. The  local  stresses  at  C,  B,  and  0  must  be  proper- 
ly provided  for  by  evident  means.  The  shear  J  =  P0,  at 
0,  also  requires  special  attention. 

This  shape  of  beam  is  often  adopted  in  practice  for  the 
working  beams  of  engines,  etc. 

The  parabolic  outlines  just  found  may  be  replaced  by 
trapezoidal  forms,  Fig.  287,  without  using  much  more  ma- 
terial, and  by  making  the  slop- 
ing plane  faces  tangent  to  the  p^T'""  ~f= 

parabolic    outline   at  points  T0  gK^ZZj1 '"'' ~~ 

and  T1}  half-way  between  0  and         "^^T^ "~" 

B,  and  C  and  B,  respectively.  It  PIS.  237. 

can  be  proved  that  they  contain  minimum  volumes,  among 
all  trapezoidal  forms  capable  of  circumscribing  the  given 
parabolic  bodies.  The  dangerous  sections  of  these  trape- 
zoidal bodies  are  at  the  tangent  points  TO  and  TV  This  is 
as  it  should  be,  (see  §  278),  remembering  that  the  subtan- 
gent  of  a  parabola  is  bisected  by  the  vertex. 

282.  I-Beam  of  Uniform  Strength. — Support  and  load  same 
as  in  the  preceding  §.  Fig.  288.  Let  the  area  of  the 


338 


MECHANICS   OF   ENGINEERING. 


Fio.  288. 

flange-sections  be  =  F  and  let  it  be  the  same  for  all  values 
of  x.  Considering  all  points  of  F  at  any  one  section  as  at 
the  same  distance  z  from  the  neutral  axis,  we  may  write 
7  =  z2F,  and  assuming  that  the  flanges  take  all  the  tension 
and  compression  while  the  (thin)  web  carries  the  shear,  the 
free  body  of  length  x  in  Fig.  288  gives  (moms,  about  n) 

P-=Pcx  ;  i.e.  £- = Pex  :  or,  since  p  is  to  be  constant, 

e  z 

z  =  [Const.],  x        .        .        .,        .    '       (1) 

i.  e.  z  must  be  made  proportional  to  x. 

Hence  the  flanges  should  be  made  straight.     Practically, 
if  they  unite  at  G,  the  web  takes  but  little  shear. 

283.  Rectang.  Section.    Height  Constant.    Two  Supports  (at  Ex- 
tremities). Single  Eccentric  Load. 
—Fig.   289.    b  and  h  are  the 
dimensions  of  the  section  at 
B.     With  BO  free  we  have 


eB 


PIG.  289. 


At  any  other  section  on  BO,  as  n,  where  the  width  is  u, 
the  variable  whose  relation  to  x  is  required,  we  have  for 
wOfree 


Equating  pK.  and  pn  we  have  u  :  b  ::  x  :  1Q 


(2) 
(3) 


FLEXURE.      BEAMS   OF   UNIFORM   STRENGTH. 


339 


That  is,  BO  must  be  wedge-shaped  with  its  edge  at  0,  ver- 
tical. 

284.  Similar  Eectang.  Sections.  Otherwise  as  Before.— Fig.  289  a, 
b  and  h  are  the  dimen- 
sions at  B;  at  any  other 
section  n,  on  BO,  the  height 
v  and  width  u,  are  the 
variables  whose  relation  to 
x  is  desired  and  by  hypoth- 
esis are  connected  by  the 
relation 


u  :  v  ::  b  :  h 


(1) 


(since  the  section  at  n  is  a  rectangle  similar  to  that  at  B). 


For  the  free  body  BO pB  = 


For  the  free  body  nO pn  =  -±-^ 

Writing  plt  =  pB  we  obtain  IQ  -r-  bh2  =  x 
put  u  =  bv  -s-  h,  from  (1) ;  whence 


•  •     (2) 

..        . '      (3) 
uv2,  in  which 

•  '    •      (4) 


which  is  the  equation  of  the  curve  (a  cubic  parabola) 
whose  abscissa  is  x  and  ordinate  l/^  v  ;  i.e.,  of  the  upper 
curve  of  the  outline  of  the  central  longitudinal  vertical 
plane  section  of  the  body  (dotted  line  BO)  which  is  sup- 
posed symmetrical  about  such  a  plane.  Similarly  the 
central  horizontal  plane  section  will  cut  out  a  curve  a 
quarter  of  which  (dotted  line  B  0)  has  an  equation 


(5) 


That  is,  the  height  and  width  must  vary  as  the  cube  root 


340  MECHANICS    OF   ENGINEERING. 

of  the  distance  from  the   support.     The  portion  CB  will 
give  corresponding  results,  referred  to  the  support  G. 

[If  the  beam  in  this  problem  is  to  have  circular  cross- 
sections,  let  the  student  treat  it  in  the  same  manner.] 

286.  Uniform  Load.  Two  End  Supports.  Rectangular  Cr. 
Sections.  Width  Constant.  — 
"Weight  of  beam  neglected. 
How  should  the  height  vary, 
the  height  and  width  at  the 
middle  being  h  and  5  ?  Fig. 
289  &.  From  symmetry  each 
reaction  =  yz  W  =  ^  wl. 
At  any  cross  section  n,  the  width  is  =  b,  (same  as  that  at 
the  middle)  and  the  height  =  v,  variable.  S  (moms.n)  =0, 
for  the  free  body  nO,  gives 

t         •     •       W 


But  for  a  beam  of  uniform  strength,  pn  is  to  be  —  pB  as 
computed  from  2  (moms.B)  =  0  for  the  free  body  .  .  BO, 
i.e  from 


_ 

2        2  w 

Hence  solve  (1)  for  pn  and  (2)  for  pB  and  equate  the  results, 
whence  t^=*L[Z*-rf]  ;  or  (^)2=(          [Z*-*2]     .     (3) 


This  relation  between  the  abscissa  a?  and  the  ordinate  y£v, 
of  the  curve  CBO,  shows  it  to  be  an  ellipse  since  eq.  (3) 
is  that  of  an  ellipse  referred  to  its  principal  diameter  and 
the  tangent  at  its  vertex  as  co-ordinate  axes. 

In  this  case  eq.  (3)  covers  the  whole  extent  of  both 
upper  and  lower  curves,  i.e.  the  complete  outline,  of  the 
curve  CBOB',  whereas  in  Figs.  286,  289,  and  289  a,  such  is 
not  the  case. 


FLEXURE.       BODIES   OF   UNIFORM   STRENGTH.      341 


287.  Cantilevers  of  Uniform  Strength,— Beams  built  in  at 
one  end,  horizontally,  and  projecting  from  the  wall  with- 
out support  at  the  other,  should  have  the  forms  given  be- 
low, for  the  given  cases  of  loading,  if  all  cross-sections  are 
to  be  Rectangular  and  the  weight  of  beam  neglected.  Sides 
of  sections  horizontal  and  vertical.  Also,  the  sections  are 
symmetrical  about  the  axis  of  the  piece,  b  and  h  are  the 
dimensions  at  the  wall.  1=  length.  No  proofs  given. 


Width    constant.  -, 
Vertical    outline 
parabolic.       Single 
end  load. 

Height  constant. «, 
Single    end  load. 
Horizontal   outline 
triangular. 

Constant  ratio  of 
height  v  to  width  u. 
Both  outlines  cu- 
bic parabolas. 


Fig.  290,  (a). 


Fig.  290,  (b). 


i     29(X 


(1) 

(2) 
(3) 


T     ' 


342  MECHANICS   OF   ENGINEERING. 

Uniform    Load. 


angular. 


Uniform  Load." 
Height  constant. 
Horiz.  outline  is 
two  parabolas  meet- 
ing at  0  (vertex) 
with  geomet.  axes 
II  to  wall. 


Fig.  291,  (6). 


Uniform  Load.*. 
Both  outlines  semi- 
cubic     parabolas.  I  Fig.  291,  (c). 

Sections    similar 
rectangles. 


(#«)'=(#*)' 


.    .(5) 

T      (6) 
?      (67 


289.  —  Beams  and  cantilevers  of  circular  cross-sections 
may  be  dealt  with  similarly,  and  the  proper  longitudinal 
outline  given,  to  constitute  them  "  bodies  of  uniform 
strength."  As  a  consequence  of  the  possession  of  this 
property,  with  loading  and  mode  of  support  of  specified 
character,  the  following  may  be  stated  ;  that  to  find  the 
equation  of  safe  loading  any  cross-section  loJiatever  may  be 
employed.  This  refers  to  tension  and  compression.  As 
regards  the  shearing  stresses  in  different  parts  of  the  beam 
the  condition  of  "  uniform  strength  "  is  not  necessarily  ob- 
tained at  the  same  time  with  that  for  normal  stress  in  the 
outer  fibres. 


DEFLECTION     OF      BEAMS     OF      UNIFORM 
STRENGTH. 

290.  Case  of  §  283,  the  double  wedge,  but  symmetrical, 
i.e.,  ?,=$,=  j£Z,  Fig.  292.     Here  we  shall  find  the  use  of  the 


BEAMS   OF    UNIFOKM   STRENGTH.  343 

£ ^pt 


EI 

form    :         (of  the  three  forms  for  the  moment  of  the  stress 

P 
couple,  see  eqs.  (5),  (6)  and  (7),-§§  229  and  231)  of  the  most 

direct  service  in  determining  the  form  of  the  elastic  curve 
OB,  which  is  symmetrical,  and  has  a  common  tangent  at 
By  with  the  curve  BC.  First  to  find  the  radius  of  curva- 
ture, p,  at  any  section  n,  we  have  for  the  free  body  nO, 
J?(moms.n—  0),  whence 


we  have  >/12  |  uk*=  %P  $  and  .'.  p-  %  %£  .  ^        (1) 

from  which  all  variables  have  disappeared  in  the  right 
hand  member  ;  i.e.,  p  is  constant,  the  same  at  all  points  of 
the  elastic  curve,  hence  the  latter  is  the  arc  of  a  circle, 
having  a  horizontal  tangent  at  B. 

To  find  the  deflection,  d,  at  B,  consider  Fig.  292,  (6j 
where  d=KB,  and  the  full  circle  of  radius  BH*-*p  is 
drawn. 

The  triangle^O.Bjs  similar  to  YOB, 
and  .-.  KB  :  OB  :  :  OB  :  TB 
But  OB=y2l,  KB=d  and  Y£=2/> 

•'•  d=  ^»  and  •'•  •  from  e(i- 


From  eq.  (4)  §233  we  note  that  for  a  beam  of  the  same 
material  but  prismatic  (parallelopipedical  in  this  case,) 
having  the  same  dimensions,  b  and  h,  at  all  sections  as  at 

1    PT?         PP 
the  middle,  deflects  an  amount  =JQ-  ^^^i         under  a 


344 


MECHANICS   OF   ENGINEERING. 


load  P  in  the  middle  of  the  span.  Hence  the  tapering 
beam  or  the  present  §  has  only  ^  the  stiffness  of  the  pris- 
matic beam,  for  the  same  6,  A,  ?,  E,  and  P. 

291.  Case  of  §281  (Parabolic  Body),  WithZl=7(),  i.e.,  Symmet- 
rical.— Fig.  293,(a).     Bequired  the  equation  of  the  neutral 


^T~     ,,1B     4-» 

^ 
T> 

**•    i  " 

•         («) 

8 t^-H* 


line  05.     For  the  free  body  nO,  J(moms.n)=0  gives  us 

27/_ !£= — y2Px       .        ,        .       (1)' 

Fig.  293,  (6),  shows  simply  the  geometrical  relations  of  the 
problem,  position  of  origin,  axes,  etc.  OnB  is  the  neutral 
line  or  elastic  curv<^  whose  equation,  and  greatest  ordinate 
d,  are  required.  (The  right  hand  member  of  eq.  (1)"  is  made 
negative  because  d?y-+dy?  is  negative,  the  curve  being  con- 
cave to  the  axis  X  in  this,  the  first  quadrant.) 

Now  if  the  beam  were  prismatic,  /,  the  "  moment  of  in- 
ertia "  of  the  cross-section  would  be  constant,  i.e.,  the  same 
for  all  values  of  x,  and  we  might  proceed  by  taking  the  x- 
anti-derivative  of  each  member  of  (1)"  and  add  a  constant ; 
but  it  is  variable  and  is 

a;T  ,    (from  eq.  3,  §281,  putting  ^=^0 


hence  (1)"  becomes 


To  put  this  into  the  form  Const,  x    ^=func.  of  (x\  we  need 


BEAMS   OF    UNIFORM   STRENGTH.  345 


3 

only  divide  through  by  x2  ,,(and  for  brevity  denote 
-L  Ebh\+  (y2l)T  by  A)  and  obtain 


We  can  now  take  the  a>anti-derivative  of  each  member,  and 
have 

A^=—y2P(^x+^)+C    ....      (2)' 

To  determine  the  constant  C,  we  utilize  the  fact  that  at  B, 
where  x=}£l,  the  slope  dy-r-dx  is  zero,  since  the  tangent 
line  is  there  horizontal,  whence  from  (2)' 


,-.  (2)'  becomes  A       -  =P[^l-x     ]     .....     (2) 


Ay=P  WJ£i.x-y3  x*]+[C'=0]    ...     (3) 

(C"=0  since  for  x=0,  y=Q).  We  may  now  find  the  deflec- 
tion d  (Fig.  293(6))  by  writing  x=  y2l  and  y  =d,  whence,  after 
restoring  the  value  of  the  constant  A, 


D73 

and  is  twice  as  great  [being=2.  _  r]*  as  if  the  girder 

4:Ebhi 

i     3 

*  See  $  233,  putting  /  =  —  bh  in  eq.  (4). 

were  parallelopipedical.     In  other  words,  the  present  girder 
is  only  half  as  stiff  as  the  prismatic  one. 

292.    Special  Problem.    (I.)     The  symmetrical  beamjn  Fig. 
294  is  of  rectangular  cross-section  and  constant  width  =  ft, 


346 


MECHANICS   OF   ENGINEERING. 


but  the  height  is  constant  only  over  the  extreme  quarter 
spans,  being—  ^=^4 h,  i.  6'»  half  the  height  li  at  mid-span. 
The  convergence  of  the  two  truncated  wedges  forming  the 
middle  quarters  of  the  beam  is  such  that  the  prolongations 


of  the  upper  and  lower  surfaces  ivould  meet  over  the  supports 
(as  should  be  the  case  to  make  A=2A,).  Neglecting  the 
weight  of  the  beam,  and  placing  a  single  load  in  middle,  it 
is  required  to  find  the  equation  for  safe  loading ;  also  the 
equations  of  the  four  elastic  curves ;  and  finally  the  deflec- 
tion. 

The  solutions  of  this  and  the  following  problem  are  left 
to  the  student,  as  exercises.  Of  course  the  beam  here 
given  is  not  one  of  uniform  strength. 

293.  Special  Problem.  (II).  Fig.  295.  Eequired  the  man- 
ner in  which  the  width  of  the  beam  must  vary,  the  height 
being  constant,  cross-sections  rectangular,  weight  of  beam 


FIG.  295. 


neglected,  to  be  a  beam  of  uniform  strength,  if  the  load  is 
uniformly  distributed  ? 


FLEXURE.      OBLIQUE  PORCBS-  347 


CHAPTER  V. 

FLEXURE  OF  PRISMATIC  BEAMS  UNDER 
OBLIQUE  FORCES. 


294,  Remarks,  By  "  oblique  forces  "  will  be  understood 
external  forces  not  perpendicular  to  tho  beam,  but  these 
external  forces  will  be  confined  to  one  plane,  called  the 
force-plane,  which  contains  the  axis  of  the  beam  and  also 
cuts  the  beam  symmetrically.  The  curvature  induced  by 
these  external  forces  will  as  before  be  considered  very 
slight,  so  that  distances  measured  along  the  beam  will  be 
treated  as  unchanged  by  the  flexure. 

It  will  be  remembered  that  in  previous  problems  the 
proof  that  the  neutral  axis  of  each  cross  section  passes 
through  its  centre  of  gravity,  rested  on  the  fact  that  when 
a  portion  of  the  beam  having  a  given  section  as  one  of  its 
bounding  surfaces  is  considered  free,  the  condition  of 
equilibrium  ^'  (compons.  ||  to  beam)=0  does  not  introduce 
any  of  the  external  forces,  since  these  in  the  problems  re- 
ferred to,  were  "|  to  the  beam ;  but  in  the  problems  of  the 
present  chapter  such  is  not  the  case,  and  hence  the  neutral 
axis  does  not  necessarily  pass  through  the  centre  of  gravity 
of  any  section,  and  in  fact  may  have  only  an  ideal,  geomet- 
rical existence,  being  sometimes  entirely  outside  of  the 
section ;  in  other  words,  the  fibres  whose  ends  are  exposed 
in  a  given  section  may  all  be  in  tension,  (or  all  in  compres- 
sion,) of  intensities  varying  with  the  distance  of  each  from 
the  neutral  axis.  It  is  much  more  convenient,  however,  to 
take  for  an  axis  of  moments  the  gravity  axis  parallel  to  the 


348  MECHANICS   OF   ENGINEERING. 

neutral   axis  instead  of  the  neutral  axis  itself,  since  this 
gravity  axis  has  always  a  known  position. 

295.  Classification  of  the  Elastic  Forces.  Shear,  Thrust,  and 
Stress-Couple.  Fig.  296.  Let  AKM\>&  one  extremity  of  a 
portion,  considered  free,  of  a  prismatic  beam,  under  oblique 
forces.  C  is  the  centre  of  gravity  of  the  section  ex- 
posed, and  GC  the  gravity  axis  ~|  to  the  force  plane  GAK. 
The  stresses  acting  on  the  elements  of  area  (each  =dF )  of 
the  section  consist  of  shears  (whose  sum=«7,  the  "total 
shear")  in  the  plane  of  the  section  and  parallel  to  the  force 
plane,  and  of  normal  stress  parallel  to  AK  and  proportion- 
al per  unit  of  area  to  the  distances  of  the  dF's  on  which 
they  act  from  the  neutral  axis  NC",  real  or  ideal  (ideal  in 
this  figure).  Imagine  the  outermost  fibre  KA,  whose  dis- 
tance from  the  gravity  axis  is=e  and  from  the  neutral  axis 


-Pi-- 


S96. 


==e  +a,  to  be  prolonged  an  amount  AA',  whose  length  by- 
some  arbitrary  scale  represents  the  normal  stress  (tension 
01  compression)  to  which  the  dF  at  A  is  subjected.  Then, 
ii  a  plane  be  passed  through  A'  and  the  neutral  axis  NC", 
tne  lengths,  such  as  mr,  parallel  to  AA',  intercepted  between 
this  plane  and  the  section  itself,  represent  the  stress-inten- 


FLEXURE.       OBLIQUE     FORCES.  349 

sities  (i.  e.,  per  unit  area)  on  the  respective  dF's.  (In  thia 
particular  figure  these  stresses  are  all  of  one  kind,  all  ten- 
sion or  all  compression  ;  but  if  the  neutral  axis  occurs 
within  the  limits  of  the  section,  they  will  be  of  opposite 
kinds  on  the  two  sides  of  NC."}  Through  C",  the  point 
determined  in  A'NC"  by  the  intercept  CC'  of  the  centre  of 
gravity,  pass  a  plane  A"M"T"  parallel  to  the  section  it- 
self ;  it  will  divide  the  stress-intensity  AA'  into  two  parts 
Pi  and  p2,  and  will  enable  us  to  express  the  stress-intensity 
mr,  on  any  dF  at  a  distance  a  from  the  gravity -axis  GC,  in 
two  parts  ;  one  part  the  same  for  all  dF's,  the  other  depen- 
dent on  2,  thus  : 

[Stress-intensity  on  any  dF]  =  PI+  — p2     .     .  (1) 
and  hence  the 
[actual  normal  stress  on  any  dF]  =  p{d  F  +  —  p2  dF    (2) 

For  example,  the  stress-intensity  on  the  fibre  at  T,  where 

2  =  —  e,,  will  be  pi  —  ^-  /%,  and  it  is  now  seen  Low  we  may 

find  the  stress  at  any  dF  when  pl  and  p^  have  been  found. 
If  the  distance  a,  between  the  neutral  and  gravity  axes  is 
desired,  we  have,  by  similar  triangles 

p2  :  e  ::  C'O'.a  whence  a  =&  .  e        .        .        .        .         (3) 

It  is  now  readily  seen,  graphically,  that  the  stresses  or  elas- 
tic forces  represented  by  the  equal  intercepts  between  the 
parallel  planes  AMT  and  A"  M"  T",  constitute  a  uniform- 
ly distributed  normal  stress,  which  will  be  called  the  "  uni- 
form thrust,"  or  simply  the  thrust  (or  pull,  as  the  case  may 

be)  of  an  intensity  =  plt  and  .•.  of  an  amount=  Cp\dF  = 

Pi  fdF  =  plF. 

It  is  also  evident  that  the  positive  intercepts  forming  the 


350 


MECHANICS   OF   ENGINEERING. 


wedge  A" A'  G'  G'  and  the  negative  intercepts  forming  the 
wedge  M"M.'G'G'  form  a  system  of  "graded  stresses" 
whose  combination  (algebraic)  with  those  of  the  "  thrust  " 
shows  the  two  sets  of  normal  stresses  to  be  equivalent  to 
the  actual  system  of  normal  stresses  represented  by  the 
small  prisms  forming  the  imaginary  solid  AMT  . .  A'M'  T'. 
It  will  be  shown  that  these  graded  stresses  constitute  a 
"  stress-couple." 

Analytically,  the  object  of  this  classification  of  the  nor- 
mal stresses  into  a  thrust  and  a  stress-couple,  may  be  made 
apparent  as  follows : 

In  dealing  with  the  free  body  KAM  Fig.  296,  we  shall 
have  occasion  to  sum  the  components,  parallel  to  the  beam, 
of  all  forces  acting  (external 
and  elastic),  also  those  ~|  to 
the  beam;  and  also  sum  their 
moments  about  some  axis  ~| 
to  the  force  plane.  Let  this 
axis  of  moments  be  GC  the 
gravity -axis  of  the  section 
(and  not  the  neutral  axis) ; 
also  take  the  axis  X  ||  to  the 
beam  and  Y~\  to  it  (and  in 
force-plane).  Let  us  see 
what  part  the  elastic  forces  Flo  297. 

will  play  in  these  three  summations.  See  Fig.  297,  which 
gives  merely  a  side  view.  Referring  to  eq.  (2)  we  see  that 


[see  eq.  (4)  §  23].     But  as  the  z's  are  measured  from  G 
gravity  axis,  z  must  be  zero.     Hence 


[The  IX  of  the  Elastic  forces]  =PlF= 


(4) 


FLEXURE.      OBLIQUE   FORCES.  351 

Also, 

[The  I  T  of  the  Elastic  forces]  =  e7==  the  shear;      .     (6) 
while  for  moments  about  G  [see  eq.  (1)] 
[The  2  (moms.G)  of  the  elastic  forces]  = 


and  hence  finally 


where  7G,  —  C  z2dF,  is  the  "moment  of  inertia  "  of 

i 

the  section  about  the  gravity  axis  G,  (not  the  neutral  axis). 
The  expression  in  (6)  may  be  called  the  moment  of  the 
stress-couple,  understanding  by  stress-couple  a  couple  to 
which  the  graded  stresses.  of  Fig.  297  are  equivalent.  That 
these  graded  stresses  are  equivalent  to  a  couple  is  shown 
by  the  fact  that  although  they  are  X  forces  they  do  not 
appear  in  eq.  4,  for  IX;  hence  the  sum  of  the  tensions 

^-f  zdF  ~|  equals  that  of  the  compressions  f^  C^zdF  "I 

in  that  set  of  normal  stresses. 

We  have  therefore  gained  these  advantages,  that,  of  the 
three  quantities  J"(lbs.),  pv  (Ibs.  per  sq.  inch),  and  p2  (Ibs. 
per  sq.  inch)  a  knowledge  of  which,  with  the  form  of  the 
section,  completely  determines  the  stresses  in  the  section, 
equations  (4),  (5),  and  (6)  contain  only  one  each,  and  hence 
algebraic  elimination  is  unnecessary  for  finding  any  one 
of  them  ;  and  that  the  axis  of  reference  of  the  moment  of 
inertia  /is  the  same  axis  of  the  section  as  was  used  in 
former  problems  in  flexure. 

Another  mode  of  stating  eqs.  (4),  (5)  and  (6)  is  this  :  The 
sum  of  the  components,  parallel  to  the  beam,  of  the  exter- 
nal forces  is  balanced  by  the  thrust  or  pull;  those  perpen- 


352  MECHANICS   OF    ENGINEERING. 

dicular  to  the  beam  are  balanced  by  the  shear;  while  the 
sum  of  the  moments  of  the  external  forces  about  the 
gravity  axis  of  the  section  is  balanced  by  that  of  the  stress- 
couple.  Notice  that  the  thrust  can  have  no^moment  about 
the  gravity  axis  referred  to. 

The  Equation  for  Safe  Loading,  then,  will  be  this : 
(a)      .     (pi±p>)  max.      "1  whichever  "] 

is     r-jr  ,    .  (7) 

greater<     J 

For  E',  see  table  in  §  251.  The  double  sign  provides  for 
the  cases  where  p{  and  p2  are  of  opposite  kinds,  one  tension 
the  other  compression.  Of  course  (pi+pz)  max  is  not  the 
same  thing  as  [pY  max.  +pz  max.].  Inmost  cases  in  prac- 
tice el=e,  and  then  the  part  (b)  of  eq.  (7)  is  unnecessary. 

295a.  Elastic  Curve  with  Oblique  Forces. — (By  elastic  curve 
is  now  meant  the  locus  of  the  centres  of  gravity  of  the  sec- 
tions.) Since  the  normal  stresses  in  a  section  differ  from 
those  occuring  under  perpendicular  forces  only  in  the  ad- 
dition of  a  uniform  thrust  (or  pull),  whose  effect  on  the 
short  lengths  (=dx)  of  fibres  between  two  consecutive  sec- 
tions V  V  and  U0V0,  Fig.  297,  is  felt  equally  by  all,  the  loca- 
tion of  the  centre  of  curvature  R,  is  not  appreciably  differ- 
ent from  what  it  would  be  as  determined  by  the  stress- 
couple  alone. 

Thus  (within  the  elastic  limit),  strains  being  proportional 
to  the  stresses  producing  them,  if  the  forces  of  the  stress- 
couple  acted  alone,  the  length  dx=G0G'  of  a  small  portion 
of  a  fibre  at  the  gravity  axis  would  remain  unchanged,  and 
the  lengthening  and  shortening  of  the  other  fibre-lengths 
between  the  two  sections  U0  V0  and  IF  V ,  originally  parallel, 
would  occasion  the  turning  of  TJ'  V  through  a  small  angle 
(relatively  to  U0  V^)  about  G',  into  the  position  which  it  oc- 
cupies in  the  figure  (297),  and  G»Rl  would  be  the  radius  of 
curvature.  But  the  effect  of  the  uniform  pull  (added  to 
that  of  the  couple)  is  to  shift  U'  V  parallel  to  itself  into 
the  position  UV,  and  hence  the  radius  of  curvature  of  the 


FLEXURE.      OBLIQUE   FORCES. 


353 


elastic  curve,  of  which  G0G  is  an  element,  is  G^R  instead 
of  GJt'.  But  the  difference  between  G0R  and  G^R'  is  very 
small,  being  the  same,  relatively,  as  the  difference  between 
GQG  and  G0G' ;  for  instance,  with  wrought-iron,  even  if  plt 
the  intensity  of  the  uniform  pull,  were  as  high  as  22,000 
Ibs.  per  sq.  in.  [see  §  203]  G0G  would  exceed  GQG'  by  only 
l/u  of  one  per  cent.  (=0.0008) ;  hence  by  using  GB'  instead 
of  GR  as  the  radius  of  curvature  p,  an  error  is  introduced 
of  so  small  an  amount  as  to  be  neglected. 

But  from  §  231,  eqs.  (6)  and  (7),  —  =  EI^  =  M,    the 

ft  CLXr 

the  sum  of  the  moments  of  the  external  forces ;  hence  for 
prismatic  beams  under  oblique  forces  we  may  still  use 


(i) 


as  one  form  for  the  J(moms.)  of  the  elastic  forces  of  the 
section  about  the  gravity-axis  ;  remembering  that  the  axis 
X  must  be  taken  parallel  to  beam. 

296.  Oblique  Cantilever  with  Terminal  Load.— Fig.  298.  Let 
1=  length.  The  "  fixing"  of  the  lower  end  of  the  beam  is 
its  only  support.  Measure  x  along  the  beam  from  0.  Let 


Fio.  299. 


n  be  the  gravity  axis  of  any  section  and  nT,  =x  sin  a,  the 
length  of  the  perpendicular  let  fall  from  n  on  the  line  of 
action  of  the  force  P  (load).  The  flexure  is  so  slight  that 
n  T  is  considered  to  be  the  same  as  before  the  load  is  al- 


,354  MECHANICS   OP   ENGINEERING. 

lowed  to  act.  [If  a  were  very  small,  however,  it  is  evident 
that  this  assumption  would  be  inadmissible,  since  then  a 
large  proportion  of  nT  would  be  due  to  the  flexure  caused 
by  the  load.] 

Consider  nO  free,  Fig.  299.  In  accordance  with  the  pre- 
ceding paragraph  (see  eqs.  (4),  (5),  and  (6))  the  elastic 
forces  of  the  section  consist  of  a  shear  J,  whose  value  may 
be  obtained  by  writing  ^'7=0 

whence  J=P  sin  « ;         .    -.  .+  •      *  ,      •     (1) 

of  a  uniform  thrust  =piF,  obtained  from  2X=Q,  viz : 

P  cos  «—  plF=Q  .'.  piF=P  cos  a. ;  .  (2) 
and  of  a  stress-couple  whose  moment  [which  we  may  write 
either  t=-,  or  El  —^  ]  is  determined  from  ,T(moms.n)=0  or 

P^—Px  sin  a=0,  or  P^=Px  sin  a      .         .      (3) 

As  to  the  strength  of  the  beam,  we  note  that  the  stress-in- 
tensity, pi,  of  the  thrust  is  the  same  in  all  sections,  from  0 
to  L  (Fig.  298),  and  that  p,,  the  stress-intensity  in  the  outer 
fibre,  (and  this  is  compression  if  e=no'  of  Fig.  299)  due  to 
the  stress-couple  is  proportional  to  x ;  hence  the  max.  of 
[/>!+#>]  will  be  in  the  lower  outer  fibre  at  L,  Fig.  298, 
where  x  is  as  great  as  possible,  =1 ;  and  will  be  a  compres- 
sion, viz. : 

.  =  P  [  cos  «  +foin  a)  e~\     ^    ^      (4) 
L     F  I      J 

.•.  the  equation  for  Safe  Loading  is 

R  =P  rcosg  j_^(sin  «)<?"]  (5\ 

L  F      ~r~] 

since  with  ei=e,  as  will  be  assumed  here,[jt>—   — #>]    max. 


FLEXURE.      OBLIQUE   FORCES. 


355 


can  not  exceed,  numerically,  [pi+p-H  max.  The  stress- 
intensity  in  the  outer  fibres  along  the  upper  edge  of  the 
beam,  being  —p\—p^  (supposing  el=e}  will  be  compressive 
at  the  upper  end  near  0,  since  there  p2  is  small,  x  being 
small ;  but  lower  down  as  x  grows  larger,  p2  increasing,  a 
section  may  be  found  (before  reaching  the  point  L)  where 
Pz=pi  and  where  consequently  the  stress  in  the  outer  fibre 
is  zero,  or  in  other  words  the  neutral  axis  of  that  section 
passes  through  the  outer  fibre.  In  any  section  above  that 
section  the  neutral  axis  is  imaginary,  i.e.,  is  altogether  out- 
side the  section,  while  below  it,  it  is  within  the  section,  but 
cannot  pass  beyond  the  gravity  axis.  Thus  in  Fig.  300,  O'L' 


FIG.  300.  FIG.  301. 

is  the  locus  of  the  positions  of  the  neutral  axis  for  successive 
sections,  while  OL  the  axis  of  the  beam  is  the  locus  of  the 
gravity  axes  (or  rather  of  the  centres  of  gravity)  of  the 
sections,  this  latter  line  forming  the  "  elastic  curve  "  un- 
der flexure.  As  already  stated,  however,  the  flexure  is  to 
be  but  slight,  and  a  must  not  be  very  small.  For  in- 
stance, if  the  deflection  of  0  from  its  position  before  flex- 
ure is  of  such  an  amount  as  to  cause  the  lever-arm  OR  of 
P  about  L  to  be  greater  by  10  per  cent,  than  its  value 
(=1  sin  a)  before  flexure,  the  value  of  p2  as  computed  from 
eq.  (3)  (with  x=T)  will  be  less  than  its  true  value  in  the 
same  proportion. 

The  deflection  of  0  from  the  tangent  at  L,  by  §  237,  Fig. 
229(a)  is  d=(P  sina)P-r-3EI,  approximately,  putting  P  sin  a 


356 


MECHANICS   OF   ENGINEERING. 


for  the  P  of  Fig.  229  ;  but  this  very  deflection  gives  to  the 
other  component,  P  cos  #,  ||  to  the  tangent  at  L,  a  lever 
arm,  and  consequent  moment,  about  the  gravity  axes  of  all 
the  sections,  whence  for  -'  (moms.L)=0  we  have,  (more  ex- 
actly than  from  eq.  (3)  when  x=l) 


(6) 


("We  have  supposed  P  replaced  by  its  components  ||  and 
"I  to  the  fixed  tangent  at  L,  see  Fig.  301).  But  even  (6) 
will  not  give  an  exact  value  for  p.2  at  L  ;  for  the  lever  arm 
of  P  cos  a,  viz.  d,  is  >(P  sin  a)P-±3EI,  on  account  of  the 
presence  and  leverage  of  P  cos  a  itself.  The  true  value  of 
d  in  this  case  may  be  obtained  by  a  method  similar  to  that 
indicated  in  the  next  paragraph. 

297.  Elastic  Curve  of  Oblique  Cantilever  with  Terminal  Load. 
More  Exact  Solution.  For  variety  place  the  cantilever  as  in 
Fig.  302,  so  that  the  deflection 
OY=  d  tends  to  decrease  the 
moment  of  P  about  the  gravity 
axis  of  any  section,  n.  AVe 
may  replace  P  by  its  X  and  Y 
components,  Fig.  303,  ||  and 
~|  respectively  to  the  fixed 
tangent  line  at  L.  The  origin, 

DJp  */WX^  o,  is  taken  at  the  free  end  of 

Fl°-  808-  Fm.303.       the  beam.     Let  a=  angle  bet- 

ween P  and  X.  For  a  free  body  On,  n  being  any  section, 
we  have  2'  (moms.n)=0 


whence 


;  P(cos  a)y — P(sin  a)  x 


(1) 


[See  eq.  (1)  §  295a].  In  this  equation  the  right  hand 
member  is  evidently  (see  fig.  303)  a  negative  quantity; 
this  is  as  it  should  be,  for  Eld^y-^dy?  is  negative,  the  curve 
being  concave  to  the  axis  X  in  the  first  quadrant.  (It 
must  be  noted  that  the  axis  X  is  always  to  be  taken  ||  to 
the  beam,  for  EId2y -t-dx?  to  represent  the  moment  of  the 
stress-couple.)  • 


FLEXURE.      OBLIQUE     FORCES.  357 

Eq.  (1)  is  not  in  proper  form  for  taking  the  a>anti-deri- 
vative  of  both  members,  since  one  term  contains  the  vari- 
able y,  an  unknown  function  of  x.  Its  integration  is  in- 
cluded in  a  more  general  case  given  in  some  works  on  cal- 
culus, but  a  special  solution  by  Prof.  Kobinson,  of  Ohio, 
is  here  subjoined  for  present  needs.* 

~\Ve  thus  obtain  as  the  equation  of  the  elastic  curve  in 
Pig.  303, 


*  ^  +  e^ir(sin«)x-(cos%l=smarer—  e^l  (2) 

In  which  e0  denotes  the  NaperianBase=2.71828,  an  abstract 
number,  and  q  for  brevity  stands  for 


To  find  the  deflection  d,  we  make  x—l  in  (2),  and  solve 
for  y;  the  result  is  d. 

The  uniform  thrust  at  L  is  piF=Pcosa  ....  (3) 
while  the  stress  intensity  p^  in  the  outer  fibre  at  L,  is  ob- 

*  Denoting  Pcoa+EIlyy  q*  and  P&ma+EIby  p*,  eq.  (l)becomea-~^=gty—ptx    .    .  (6) 
Differentiate  (6)  then  —^=?9—^-  —p".  Differentiate  again  :  whence-  ~=g-*-^|p  .  (7) 

Letting^-    • 

See  (7)  <j?L=q*u,  which'  mult-  by  2<*«'  ^es  -^  2  du  d*u=2g*vdu 

•  '•    (    dx  constant    J,  ^      /2dudM=2?»       /  vdit+C  .:'j^=q*v*+C.    whe 


.     qx  qx  r  n 

Square  each   side  of  (9);  then    C'e    —We    w+«»=u*+-V,    .'.  u=K  ?«    —  _5_ 
n  n  9  n       ZC's* 

-.    (see  eqs.  6  and  8)  ffV-J*»-X  ^t™*™  *       ^^lldatlng  the 


358  MECHANICS   OF   ENGINEERING. 

tained  from  the  moment  equation  for  the  free  body  0  L 
viz:  &L=P(<$ma}l—  P(cosa)d     .....  (4) 


in  which  e=  distance  of  outer  fibre  from  the  gravity  axis. 
The  equation  for  safe  loading  is  written  out  by  placing 
the  values  of  plt  p^>  and  d,  as  derived  from  equations  (2)> 
(3),  and  (4)  in  the  expression 


To  solve  the  resulting  equation  for  P,  in  case  that  is  the 
unknown  quantity,  can  only  be  accomplished  by  successive 
assumptions  and  approximations,  since  it  occurs  trans- 
cendentally. 

In  case  a  horizontal  tension-member  of  a  bridge-truss  is 
subjected  to  a  longitudinal  tension  P'  (due  to  its  position 
in  the  truss  and  the  load  upon  the  latter)  and  at  the  same 
time  receives  a  vertical  pressure—  P"  at  the  middle,  each 
half  will  be  bent  in  the  same  manner  as  the  cantilever  in 
Fig.  302  ;  ^P"  corresponding  to  P  sin  a,  and  P'to  Pcoso. 

~fl          qx       —yx 
constant  factors  we  now  write  y  =  -^x-^me   —  ne        .    .    the  equation  required  .   (10) 


To  determine  the  constants  m  and  n(m=  C'-t-2q*;  n=  C-t-2C'q*)  we  first  find  dy-t-dx,  i.e. 
by  differentiating  (10)  -^=~^+Qmen  +V™n        ....    (11)          z=0  for  y=0 
.'.  (10)  gives     .    .    .    0=0+m«n—  n«n  =0    i.e.  m—  »=0  .-.  m=»   .     (12)    Also  for  x=l 

dv  n1  Ql    —yl~\ 

^-=0.'.  (11)   gives  0=^+0     wi«n  -fn«n         .    .    (13);   .'.with  n=m  we  have  m=n= 
•    •    -(14).    The  equation  of  the  curve,  then,  substituting  (14)  in 


(10)  isy=x-    -  -    —  __        .  (15)    .-.,  Substituting  for  p  and  q  we 

4+«n 

nave,  (as  in  §897)      /&jf     V*"n  aisina-ycosa       =sin«       **-*"** 


FLEXURE.       OBLIQUE   FORCES. 


359 


298.  Inclined  Beam  with  Hinge  at  One  End.— Fig.  304.  Let 
t  —  €i.  Bequired  the  equation  for  safe  loading  ;  also  the 
ru3ximum  shear,  there  being  but  one  load,  P,  and  that  in 
tirj  middle.  The  vertical  wall  being  smooth,  its  reaction, 


B,  Ai  0  is  horizontal,  while  that  of  the  hinge-pin  being  un- 
known, both  in  amount  and  direction,  is  best  replaced  by 
its  horizontal  and  vertical  components  H0  and  F0,  unknown 
in  amount  only.  Supposing  the  flexure  slight,  we  find 
these  external  forces  in  the  same  manner  as  in  Prob.  1  § 
37,  by  considering  the  whole  beam  free,  and  obtain 


=?  cota  ;  H0  also  =  £  cota  ;  F0  =  P 


(1) 


For  any  section  n  between   0  and  B,  we  have,  from  the 
free  body  nO,  Fig.  305, 


uniform  thrust  =  j^F  =  H  cos  a 
and  from  2*  (moms.n)  =  0, 


•    (2) 


=Hx  sin  a 


(3) 


and  the  shear  =  J  =  5"  sin  a  =  %  P  cos  a  (4) 

The  max.  (/>i+ft)  to  be  found  on  OB  is  .-.  close  above  B, 
where  x  =  ^  I,  and  is 


360  MECHANICS  OF   ENGINEERING. 

=  p  co 


In  examining  sections  on  CB  let  the  free  body  be  Cn't 
Fig.  306.     Then  from  2  (longitud.  comps.)  =  0 

(the  thrust=)  p^F  =  F0  sin  a  +  H0  cos  a  (6)' 

i.e.                    plF=P[ain  a  +  }4  cos  a  cot  a]  (6) 
while,  from  2'(moms.n')  =  0, 

sin  a  (7)' 


(7) 


Hence  (p}  +  P*)  for  sections  on  CB  is  greatest  when  x' 
is  greatest,  which  is  when  x'  =  ]/2  I,  x'  being  limited  be- 
tween x'  =  0  and  x'  =  y2  I,  and  is 


max.  on  <7£=P  cos  «tan  g+      Cot  a+^  (8) 


which  is  evidently  greater  than  the  max.  (pi 

see  eq.  (5).     Hence  the  equation  for  safe  loading  is 


B'  =  P  cos  aL+X  ....    (9) 


in  which  R'  is  the  safe  normal  stress,  per  square  unit,  for 
the  material. 

The  shear,  J,  anywhere  on  CBttxom2  (transverse comp.) 
=0  in  Fig.  306,  is 

J=  PO  cos  a  —  HO  sin  a  =  tf  P  cos  a       .         .  (10) 


.301 


FLEXURE.       OBLIQUE   FORCES. 


As  showing  graphically  all  the  results  found,  moment, 
thrust,  and  shear  diagrams  are  drawn 
in  Fig.  307,  and  also  a  diagram  whose 
ordinates  represent  the  variation  of 
(p^p^  along  the  beam.  Each  ordi- 
nate  is  placed  vertically  under  the 
gravity  axis  of  the  section  to  which  it 
refers. 

299.  Numerical  Example  of  the  Forego- 
ing.—  Fig.  308.  Let  the  beam  be  of 
wrought  iron,  the  load  P  =  1,800  Ibs., 
hanging  from  the  middle.  Cross  sec- 
tion rectangular  2  in.  by  1  in.,  the  2 
in.  being  parallel  to  the  force-plane. 
Required  the  max.  normal  stress  in 
-any  outer  fibre  ;  also  the  max.  total 
shear. 

This  max.  stress  -intensity  will  be  in  FlG.  307. 

the  outer  fibres  in  the  section  just  below  B  and  on  the 
upper  side,  according  to  §  298,  and  is  given  by  eq.  (8)  of 
that  article  ;  in  which,  see  Fig.  308,  we  must  substitute 
{inch-pound-second-system)  P  =  1,800  Ibs.;  F  —  2  sq.  in.; 
I  =  ^120*  +  122  =  120.6  in.;  e  =  1  in.,  /  =  |2  W  =  fi  =  2A 
biquad.  inches  ;  cot  a  =  1  ;  cos  a  =  .0996  ;  and  tan  a  =  10. 


,.  ma,. 


FIG.  808.  Fio.  309.  FIG.  310. 

9000  Ibs.  per  sq.  inch,  very  nearly,  compression.    This  is  w 


3G2  MECHANICS   OP   ENGINEERING. 

the  upper  outer  fibre  close  under  B.  In  the  lower  outer  fibre 
just  under  B  we  have  a  tension  =  p.2  —  PI  =  7,200  Ibs.  per 
sq.  in.  (It  is  here  supposed  that  the  beam  is  secure  against 
yielding  sideways.) 

300.  Strength  of  Hooks.—  An  ordinary  hook,  see  Fig.  309, 
may  be  treated  as  follows  :  The  load  being  =  P,  if  we 
make  a  horizontal  section  at  AB,  whose  gravity  axis  g  is 
the  one,  of  all  sections,  furthest  removed  from  the  line  of 
action  of  P,  and  consider  the  portion  C  free,  we  have  the 
shear  =  J  =  zero  .  .  .  .  .  •  .  (1) 

the  uniform  pull  =  p,F=P      .        .        .     (2) 

while  the  moment  of  the  stress-couple,  from  2  (moms.g)  = 
0,is 

&*=Pa  (3) 

For  safe  loading  p±  -f  p2  must  =  R',  i.e. 


Tt  is  here  assumed  that  e  =  el}  and  that  the  maximum 
occurs  at  AB. 


301  Crane.  —  As  an  exercise  let  the  student  investigate  the 
strength  of  a  crane,  such  as  is  shown  in  Fig.  310. 


FLEXURE.   LONG  COLUMNS. 


CHAPTER  VL 


FLEXURE   OF   "  LONG  COLUMNS." 


302.  Definitions. — By  "  long  column  "  is  meant  a  straight 
beam,  usually  prismatic,  which  is  acted  on  by  two  com- 
pressive  forces,  one  at  each  extremity,  and  whose   length 
is  so  great  compared  with  its  diameter  that  it  gives  way 
(or  "  fails  ")  by  buckling  sideways,  i.e.  by  flexure,  instead 
of  by  crushing  or  splitting  like  a  short  block  (see  §  200). 
The  pillars  or  columns  used  in  buildings,  the  compression 
members   of  bridge-trusses  and  roofs,  the  "  bents  "  of  a 
trestle  work,  and  the  piston-rods  and  connecting-rods 'of 
steam-engines,  are  the  principal  practical  examples  of  long 
columns.     That  they  should  be  weaker  than  short  blocks 
of  the  same  material  and  cross-section  is  quite  evident,  but 
their  theoretical  treatment  is  much   less   satisfactory  than 
in  other  cases  of  flexure,  experiment  being  very  largely 
relied  on  not  only  to  determine   the  physical    constants 
which  theory  introduces  in  the  formulae  referring  to  them, 
but  even  to  modify  the  algebraic  form  of  those  formulae, 
thus  rendering  them  to  a  certain  extent  empirical. 

303.  End  Conditions. — The  strength  of  a  column  is  largely 
dependent  on   whether  the  ends  are  free  to  turn,   or  are 
fixed  and  thus  incapable  of  turning.     The  former  condi- 
tion is  attained  by  rounding  the  ends,  or  providing  them 
with  hinges  or  ball-and-socket-joints  ;  the  latter  by  facing 
off  each  end  to  an  accurate  plane  surface,  the  bearing  on 
which  it  rests  being  plane  also,  and  incapable  of  turning. 


364 


MECHANICS    OF   ENGINEERING. 


In  the  former  condition  the  column  is  spoken  of  as  having 
round  ends ;  Fig.  311,  (a) ;  in  the  latter  as  having  fixed  ends, 
{or  flat  bases ;  or  square  ends),  Tig.  311,  (Z>). 


FIG.  312. 

Sometimes  a  column  is  fixed  at  one  end  while  the  other 
end  is  not  only  round  but  incapable  of  lateral  deviation  from 
the  tangent  line  of  the  other  extremity ;  this  state  of  end 
conditions  is  often  spoken  of  as  "Pin  and  Square,"  Fig. 
311,  (c). 

If  the  rounding  of  the  ends  is  produced  by  a  hinge  or 
*'  pin  joint,"  Fig.  312,  both  pins  lying  in  the  same  plane 
and  having  immovable  bearings  at  their  extremities,  the 
column  is  to  be  considered  as  round-ended  as  regards  flex- 
ure in  the  plane  ~|  to  the  pins,  but  as  square-ended  as  re^ 
gards  flexure  in  the  plane  containing  the  axes  of  the  pins. 

The  "  moment  of  inertia  "  of  the  section  of  a  column  will 
be  understood  to  be  referred  to  a  gravity  axis  of  the  sec- 
tion which  is  ~|  to  the  plane  of  flexure  (and  this  corres- 
ponds to  the  "  force-plane  "  spoken  of  in  previous  chap- 
ters), or  plane  of  the  axis  of  column  when  bent. 

303«.  Euler's  Formula. — Taking  the  case  of  a  round-ended 
column,  Fig.  313  (a),  assume  the  middle  of  the  length  as 
an  origin,  with  the  axis  X  tangent  to  the  elastic  curve  at 
that  point.  The  flexure  being  slight,  we  may  use  the  form 
-s-dx1  for  the  moment  of  the  stress-couple  in  any 


FLEXURE.   LONG  COLUMNS. 


365 


(a 


FIG.  314. 


section  n,  remembering  that  with  this  notation  the  axis  X 
must  be  ||  to  the  beam,  as  in  the  figure  (313).  Considering 
the  free  body  nC,  Fig.  313  (6),  we  note  that  the  shear  is 
zero,  that  the  uniform  thrust  =P,  and  that  ,T(nioms.n)=0 
gives  (a  being  the  deflection  at  0) 


(1) 


Multiplying  each  side  by  dy  we  have 

El 

-T-i  dycPy=Pady — Py  dy        .        .        (2) 
dx 

Since  this  equation  is  true  for  the  y,  dx,  dy,  and  (Py  of  any 
element  of  arc  of  the  elastic  curve,  we  may  suppose  it 
written  out  for  each  element  from  0  where  y=0,  and<%=0, 
up  to  any  element,  (where  dy=dy  and  y=y)  (see  Fig.  314) 
and  then  write  the  sum  of  the  left  hand  members  equal  to 
that  of  the  right  hand  members,  remembering  that,  since 
dx  is  assumed  constant,  1-t-dx2  is  a  common  factor  on  the 
left.  In  other  words,  integrate  between  0  and  any  point 
of  the  curve,  n.  That  is, 


f1™2      J  e/o  e/o 

ajr    «/  dy-0 

The  product  dy  d*y  has  been  written  (dy}d(dy\  (for  d?y  is 


3G()  MECHANICS   OF    ENGINEERING. 

the  differential  or  increment  of  dy]  and  is  of  a  form  like 
xdx,  or  ydy.     Performing  the  integration  we  have 


which  is  in  a  form  applicable  to  any  point  of  the  curve, 
and  contains  the  variables  x  and  y  and  their  increments 
dx  and  dy.  In  order  to  separate  the  variables,  solve  for  dx, 
and  we  have 


dx 


=  IEI 

VI 


.,a?=±-v/--  (vers.  sin—  x  .       -.        .    (6) 


.e 

(6)  is  the  equation  of  the  elastic  curve  DO  C,  Fig.  313  (a), 
and  contains  the  deflection  a.  If  P  and  a  are  both  given, 
y  can  be  computed  for  a  given  x,  and  vice  versa,  and  thus 
the  curve  traced  out,  but  we  would  naturally  suppose  a  to 
depend  on  P,  for  in  eq.  (6)  when  x=  y2l,  y  should  =a.  Mak- 
ing these  substitutions  we  obtain 

y2l=  JM  (vers.  sin  —  1.00)  ;  i.e.,  */2l=  ^    \      (7) 

Since  a  has  vanished  from  eq.  (7)  the  value  for  P  ob- 
tained from  this  equation,  viz.: 

PO=EI^  .     .  .  ./  .;  ...    (8) 

is  independent  of  a,  and 

is  ,'.  to  be  regarded  as  that  force  (at  each  end  of  the  round- 
ended  column  in  Fig.  313)  which  will  hold  the  column  at 
any  small  deflection  at  which  it  may  previously  have  been 

set. 


FLEXURE.   LONG  COLUMNS. 


367 


In  other  words,  if  the  force  is  less  than  P0  no  flexure  at 
all  will  be  produced,  and  hence  P0  is  sometimes  called  the 
force  producing  "  incipient  flexure."  [This  is  roughly  ver- 
ified by  exerting  a  downward  pressure  with  the  hand  on 
the  upper  end  of  the  flexible  rod  (a  T-square-blade  for  in- 
stance) placed  vertically  on  the  floor  of  a  room  ;  the  pres- 
sure must  reach  a  definite  value  before  a  decided  buckling 
takes  place,  and  then  a  very  slight  increase  of  pressure  oc- 
casions a  large  increase  of  deflection.] 

It  is  also  evident  that  a  force  slightly  greater  than  P0 
would  very  largely  increase  the  deflection,  thus  gaining  for 
itself  so  great  a  lever  arm  about  the  middle  section  as  to 
cause  rupture.  For  this  reason  eq.  (8)  may  be  looked 
upon  as  giving  the  Breaking  Load  of  a  column  with  round 
ends,  and  is  called  Euler's  formula. 

Referring  now  to  Fig.  311,  it  will  be  seen  that  if  the  three 
parts  into  which  the  flat-ended  column  is  di-  , 

vided  by  its  two  points  of  inflection  A  and  B 
are  considered  free,  individually,  in  Fig.  315, 
the  forces  acting  will  be  as  there  shown,  viz.: 
At  the  points  of  inflection  there  is  no  stress- 
couple,  and  no  shear,  but  only  a  thrust,  =P> 
and  hence  the  portion  AB  is  in  the  condition 
of  a  round-ended  column.  Also,  the  tangents 
to  the  elastic  curves  at  0  and  C  being  pre- 
served vertical  by  the  f  rictionless  guide-blocks 
and  guides  (which  are  introduced  here  simply 
as  a  theoretical  method  of  preventing  the  ends 
from  turning,  but  do  not  interfere  with  verti- 
cal freedom)  OA  is  in  the  same  state  of  flex- 
ure as  half  of  AB  and  under  the  same  forces. 
Hence  the  length  AB  must  =  one  half  the 
total  length  I  of  the  flat-ended  column.  In 
other  words,  the  breaking  load  of  a  round- 
ended  column  of  length  =^l,  is  the  same  as 
that  of  a  flat-ended  column  of  length  =1. 
Hence  for  the  I  of  eq.  (8)  write  l/2l  and  we 
have  as  the  breaking  load  of  a  column  with 
flat-ends  and  of  length  =1. 


PIG.  315. 


368  MECHANICS   OF   ENGINEERING. 


(9) 


Similar  reasoning,  applied  to  the  "  pin-and-square  " 
mode  of  support  (in  Fig.  311)  where  the  points  of  inflec- 
tion are  at  B,  approximately  ^  /  from  (7,  and  at  the 
extremity  0  itself,  calls  for  the  substitution  of  2/^  I  for  I  in 
eq.  (8),  and  hence  the  breaking  load  of  a  "pin-and-square  " 
column,  of  length  =  I,  is 

P,=|  EI*     .,     '  .        .        (10) 

Comparing  eqs.  (8),  (9),  and  (10),  and  calling  the  value  of 
PI  (flat-ends)  unity,  we  derive  the  following  statement  : 
The  breaking  loads  of  a  given  column  are  as  the  numbers 

f        1        I  9/16  I          ^          i    according  to  the 

I  flat-ends  \  pin-and-square  \  round-ends  \    mode  of  support. 

These  ratios  are  approximately  verified  in  practice. 

Euler's  Formula  [i.e.,  eq.  (8)  and  those  derived  from  it, 
(9)  and  (10)]  when  considered  as  giving  the  breaking  load 
is  peculiar  in  this  respect,  that  it  contains  no  reference  to 
the  stress  per  unit  of  area  necessary  to  rupture  the  material 
of  the  column,  but  merely  assumes  that  the  load  producing 
"  incipient  flexure  ",  i.e.,  which  produces  any  bending  at 
all,  will  eventually  break  the  beam  because  of  the  greater 
and  greater  lever  arm  thus  gained  for  itself.  In  the  canti- 
lever of  Fig.  241  the  bending  of  the  beam  does  not  sensibly 
affect  the  lever-arm  of  the  load  about  the  wall-section,  but 
with  a  column,  the  lever-arm  of  the  load  about  the  mid- 
section  is  almost  entirely  due  to  the  deflection  produced. 

304.  Example.  Euler's  formula  is  only  approximately 
verified  by  experiment.  As  an  example  of  its  use  when 
considered  as  giving  the  force  producing  "  incipient  flex- 
ure "  it  will  now  be  applied  in  the  case  of  a  steel  T-square- 
blade  whose  ends  are  free  to  turn.  Hence  we  use  the 
round-end  formula  eq.  (8)  of  §303,  with  the  modulus  of 
elasticity  .£7=30,000,000  Ibs.  per  sq.  inch.  The  dimensions 


FLEXURE.   LONG  COLUMNS.  369 

are  as  follows  :  the  length  I  =  30  in.,  thickness  =  i  of  an 
inch,  and  width  =  2  inches.  The  moment  of  inertia,  Ir 
about  a  gravity  axis  of  the  section  ||  to  the  width  (the 
plane  of  bending  being  ||  to  the  thickness)  is  (§247) 


/.  ,  with  TT  =  22  -5-  7, 


--     30,000,000     222        1    _ 

72-  •  go0-2.03  Ibs. 


Experiment  showed  that  the  force,  a  very  small  addition 
to  which  caused  a  large  increase  of  deflection  or  side-buck- 
ling, was  about  2  Ibs. 

305.  Hodgkinson's  Formulae  for  Columns.  —  The  principal 
practical  use  of  Euler's  formula  was  to  furnish  a  general 
form  of  expression  for  breaking  load,  to  Eaton  Hodgkin- 
son,  who  experimented  in  England  in  1840  upon  columns 
of  iron  and  timber. 

According  to  Euler's  formula  we  have  for  cylindrical 

columns,  /being  =%  KT*  =  ~  nd*  (§247), 

CVx 

for  flat-ends        ..         ,  Pl  =1  ET?  .  f 

16  t 

i.e.,  proportional  to  the  fourth  power  of  the  diameter,  and 
inversely  as  the  square  of  the  length.  But  Hodgkinson's 
experiments  gave  for  wrought-iron  cylinders 

J3.55  /T3-* 

PI  =  (const.)  x  „  —  ;  and  for  cast  iron  P!=(const.)x  j^j— 

Again,  for  a  square  column,  whose  side  =  b,  Euler's  for- 
mula would  give 


whilo  Hodgkinson  found  for  square  pillars  of  wood 


370  MECHANICS   OF   ENGINEERING. 

Hence  in  the  case  of  wood  these  experiments  indicated  the 
same  powers  for  b  and  I  as  Euler's  formula,  but  with  a  dif- 
ferent constant  factor  ;  while  for  cast  and  wrought  iron 
the  powers  differ  slightly  from  those  of  Euler. 

Hodgkinson's  formulae  are  as  follows,  and  evidently 
not  homogeneous  ;  the  prescribed  units  should  .-.  be  care- 
fully followed,  d  denotes  the  diameter  of  the  cylindrical 
columns,  b  the  side  of  square  columns,  1=  length. 

(  For  solid  cylindrical  cast  iron  columns,  flat-ends  ; 
-<  Breaking  load  in  tons  )       A  A  n  a      ,  7  .     .     ,      Ni55     n  .    .,  \  7 
(     of  2,240  Ibs.  each     |  =44.16  X  (^  in  inches)    +  (Zmft.) 

.  (  For  solid  cylindrical  wrought  iron  columns,  flat-ends  ; 
-<  Breaking  load  in  tons  )      -,04      ,-,  •    .     -,      \3-M     n  .     ,,  N2 
(     of  2,240  Ibs.  each    |  -134  X  (dm  inches)     +  (I  in  ft.) 

{For  solid  square  columns  of  dry  oak,  flat-ends  ; 


["For  solid  square  columns  of  dry  fir,  flat-ends  ; 


Hodgkinson  found  that  when  the  mode  of  support  was 
"  pin-and-square,"  the  breaking  load  was  about  ^  as 
great  ;  and  when  the  ends  were  rounded,  about  ^  as  great 
as  with  flat  ends.  These  ratios  differ  somewhat  from  the 
theoretical  ones  mentioned  in  §303,  just  after  eq.  (10.) 

Experiment  shows  that,  strictly  speaking,  pin  ends  are 
not  equivalent  to  round  ends,  but  furnish  additional 
strength  ;  for  the  friction  of  the  pins  in  their  bearings 
hinders  the  turning  of  the  ends  somewhat.  As  the  lengths 
become  smaller  the  value  of  the  breaking  load  in  Hodg- 
kinson's formulae  increases  rapidly,  until  it  becomes  larger 
than  would  be  obtained  by  using  the  formula  for  the 
crushing  resistance  of  a  short  block  (§201)  viz.,  FC,  i.e., 
the  sectional  area  X  the  crushing  resistance  per  unit  of 
area. 

In  such  a  case  the  pillar  is  called  a  short  column,  or  "  short 
block,"  and  the  value  FC  is  to  be  taken  as  the  breaking 


FLEXURE.   LONG  COLUMNS.  371 

load.  This  distinction  is  necessary  in  using  Hodgkinson's 
formulae ;  i.e.,  the  breaking  load  is  the  smaller  of  the  two 
values,  FC  and  that  obtained  by  Hodgkinson's  rule. 

In  present  practice  Hodgkinson's  formulae  are  not  often 
used  except  for  hollow  cylindrical  iron  columns,  for  which 
with  d2  and  d{  as  the  external  and  internal  diameters,  we 
have  for  flat-ends 

Breaking  load  in  tons  )  _«  _,  (c?2  in  in.)355  —  (d,  in  in.)355 

of  2,240  Ibs.  each     \~  ~(i  fr  feet)- 

in  which  the  const.  =  44.16  for  cast  iron,  and  134  for 
wrought,  while  n  =  1.7  for  cast-iron  and  =  2  for  wrought. 

306.  Examples  of  Hodgkinson's  Formulae. — Example  1.  Ee- 
quired  the  breaking  weight  of  a  wrought-iron  pipe  used 
as  a  long  column,  having  a  length  of  12  feet,  an  internal 
diameter  of  3.  in.,  and  an  external  diameter  of  3^  inches, 
the  ends  having  well  fitted  flat  bases. 

If  we  had  regard  simply  to  the  sectional  area  of  metal, 
which  is  F  =  1.22  sq.  inches,  and  treated  the  column  as  a 
short  block  (or  short  column)  we  should  have  for  its  com- 
pressive  load  at  the  elastic  limit  (see  table  §203)  P"=FC" 
=  1.22  x  24,000=29,280  Ibs.  and  the  safe  load  P1  may  be 
taken  at  16,000  Ibs. 

But  by  the  last  formula  of  the  preceding  article  we  have 

Breaking  load   in   )  _1%nx  (3.25)355-  3s-55  _  lg  07 
tons  of  2,240  Ibs.  each  f  iff 

i.e.=  15.07  x  2240=  33,768  Ibs. 

Detail,     [log.  3.25]  x  3.55=  0.511883x3.55=  1.817184 ; 

[log.  3.00]  x 3.55=0.477,121x3.55=1.693,779  ; 

and  the  corresponding  numbers  are  65.6  and  49.4 ;  their 
difference  =  16.2,  hence 

Br.  load  in  long  tons  =  134*|6'2= 15.072  long  tons. 

Irxrx. 

=33,768  Ibs. 


372  MECHANICS  OF   ENGINEERING. 

With  a  "  factor  of  safety  "  (see  §205)  of  four,  we  have,  as 
the  safe  load,  p'  =  8,442  Ibs.  This  being  less  than  the 
16000  Ibs.  obtained  from  the  "  short  block  "  formula,should 
be  adopted. 

If  the  ends  were  rounded  the  safe  load  would  be  one- 
third  of  this  i.e.,  would  be  2,814  Ibs  ;  while  with  pin-and- 
square  end-conditions,  we  should  use  one-half,  or  4,221  Ibs. 

EXAMPLE  2.  Kequired  the  necessary  diameter  to  be 
given  a  solid  cylindrical  cast-iron  pillar  with  flat  ends,  that 
its  safe  load  may  be  13,440  Ibs.  taking  6  as  a  factor  of 
safety.  Let  d  =  the  unknown  diameter.  Using  the  proper 
formula  in  §  305,  and  hence  expressing  the  breaking  load, 
which  is  to  be  six  times  the  given  safe  load,  in  long  tons 
we  have  (the  length  of  column  being  16  ft.) 

13440  x  6^  44.16  (d  in  inches)3-55  (1) 


(2) 


or  log.rf=3-ij[log.  36+1.7xlog.  16-log.  44.16]      .      .       (3) 
.-.  log.d=3-y  1.958278]  =0.551627  .-.  d  =  3.56  ins. 

This  result  is  for  flat  ends.  If  the  ends  were  rounded, 
we  should  obtain  d  =4.85  inches. 

307.  Rankine's  Formula  for  Columns.  —  The  formula  of  this 
name  (some  times  called  Gordon's,  in  some  of  its  forms)  has 
a  somewhat  more  rational  basis  than  Euler's,  in  that  it  in- 
troduces the  maximum  normal  stress  in  the  outer  fibre  and 
is  applicable  to  a  column  or  block  of  any  length,  but  still 
contains  assumptions  not  strictly  borne  out  in  theory,  thus 
introducing  some  co-efficients  requiring  experimental  de- 
termination. It  may  be  developed  as  follows  : 

Since  in  the  flat-ended  column  in  Fig.  315  the  middle 
portion  AB,  between  the  inflection  points  A  and  B,  is 
acted  on  at  each  end  by  a  thrust  =  P,  not  accompanied  by 
any  shear  or  stress-couple,  it  will  be  simpler  to  treat  tha* 


FLEXUKE.   LONG  COLUMNS. 


373 


portion  alone  Fig.  316,  (a),  since  the  thrust  and  stress- 
couple  induced  in  the  section  at 
R,  the  middle  of  AB,  will  be  equal 
to  those  at  the  flat  ends,  0  and  G, 
in  Fig.  315.  Let  a  denote  the  de- 
flection of  R  from  the  straight  line 
AB.  Now  consider  the  portion 
AR  as  a  free  body  in  Fig.  316,  (&), 
putting  in  the  elastic  forces  of  the 
section  at  R,  which  may  be  clas- 
sified into  a  uniform  thrust  = 
,  and  a  stress  couple  of  moment 


FIG.  316. 


(see  §  294). 


(The  shear   is  evidently  zero,  from 

S  (hor  coinps.)  =  0).  Here  pl  denotes  the  uniform  pres- 
sure (per  unit  of  area),  due  to  the  uniform  thrust,  and  p3 
the  pressure  or  tension  (per  unit  of  area),  in  the  elastic 
forces  constituting  the  stress-couple,  on  the  outermost 
element  of  area,  at  a  distance  e  from  the  gravity  axis  (~| 
to  plane  of  flexure)  of  the  section.  F  is  the  total  area  of 
the  section.  /  is  the  moment  of  inertia  about  the  said 
gravity  axis,  g 

I  (vert,  comps.)  =  0  gives  P  =  p,F        .        .      (1) 
J(moms.g)  =  Ogives  Pa  =&L    ....    (2) 

For  any  section,  n,  between  A  and  R,  we  would  evidently 
have  the  same  pt  as  at  R,  but  a  smaller  p2,  since  Py  <  Pa 
while  e,  /,  and  F,  do  not  change,  the  column  being  pris- 
matic. Hence  the  max.  (Pi+p<t)  is  on  the  concave  edge  at 
R  and  for  safety  should  be  no  more  than  C  -5-  n,  where  C 
is  the  Modulus  of  Crushing  (§  201)  and  n  is  a  "  factor  of 
safety."  Solving  (1)  and  (2)  for  pl  and#j,  and  putting  their 
sum  =  C  -r-  n,  we  have 


P,Pae     C 


(3) 


We  might  now  solve  for  P  and  call  it  the  safe  load,  but  it 


374  MECHANICS   OF   ENGINEERING. 

is  customary  to  present  the  formula  in  a  form  for  giving 
the  breaking  load,  the  factor  of  safety  being  applied  after- 
ward. Hence  we  shall  make  n  =  1,  and  solve  for  P,  call- 
ing it  then  the  breaking  load.  Now  the  deflection  a  is  un- 
known, but  may  be  expressed  approximately,  as  follows, 
in  terms  of  e  and  I. 

Suppose   two   columns  of       lengths   =   I'  and  I",  each 

El'     TJ  T 
bearing  its  safe  load.     Then  at  the  point  R,  —  -...G—j  i.e., 

E'e'  =  p'  p2'.     Considering  the  curve  AB  as  a  circular  arc 

we  have  (see  §  290)  a'  =  ln  -5-  32  /,  i.e.  a'  =/J^f  .  l-f  ;  and 

e 


» 

similarly  for  the  other  column,  a"  '=  -^2        .  —  .      If   the 

e 


columns  are  of  the  same  material  E'  =  E",  and  if  each  is 
bearing  its  safe  load  we  may  assume  p2'  =  p2"  nearly,  in 
which  case  the  term  p2"  -i-  E"  =  pzf  -j-  E'  ',  and  we  may 
say  that  the  deflection  a,  under  safe  load,  is  proportional 
to  (length)2  -T-  e,  approximately,  i.  e.,  that  ae  —  fil2,  where 
f}  is  a  constant  (an  abstract  number  also)  dependent  on 
experiment  and  different  for  different  materials,  and  I  the 
full  length.  We  may  also  write,  for  convenience,  I  =  Fit?, 
k  being  the  radius  of  gyration  (see  §  85).  Hence,  finally, 
we  have  from  eq.  (3) 

Breaking    load  )       p        FC 
for  flat  ends        }  =^1==     ~  V 


This  is  known  as  Rankine's  formula. 

By  the  same  reasoning  as  in  §  303,  for  a  round-ended 
column  we  substitute  2  I  for  I ;  for  a  pin-and-square  col- 
umn -t  I  for  I ;  and  .*.  obtain 

^Breaking  load  ) p  _      FC  ,~\ 

for  a  round-ended  column  j  ~    °~~     ~~j|r 

P 

Breaking  load  for  )  =p  _       FC  /fij 

a  pin-and-square  column  (     "    a  ~      p"      * 


FLEXUKE.   LONG  COLUMNS. 


375 


These  formulae,  (4),  (5),  and  (6),  unlike  Hodgkinson's, 
are  of  homogeneous  form.  Any  convenient  system  of  units 
may  therefore  be  used  in  them. 

Rankine  gives  the  following  values  for  G  and  yff,  to  be 
used  in  these  formulae.  These  are  based  on  Hodgkinson's 
experiments. 


CAST  IRON. 

WR'T  IRON. 

TIMBER. 

C  in  Ibs.  per  sq.  in. 

80,000 

36,000 

7,200 

,9  (abstract  number) 

1 

poo" 

1 
36,000 

1 
3,000 

If  these  numerical  values  of  C  are  used  F  must  be  ex- 
pressed in  Sq.  Inches  and  P  in  Pounds.  Rankine  recom- 
mends 4  as  a  factor  of  safety  for  iron  in  quiescent  struct- 
ures, 5  under  moving  loads  ;  10  for  timber.  The  N.  J. 
Iron  &  Steel  Co.  use  Rankine's  formula  for  their  wrought 
iron  rolled  beams,  when  used  as  columns,  with  a  factor  of 
safety  of  4^. 

308.  Examples,  Using  Rankine's  Formula.  —  EXAMPLE  1.  — 
Take  the  same  data  for  a  wrought  iron  pipe  used  as  a 
column,  as  in  example  1,  §  306  ;  i.e.,  Z=12  ft.=144  inches, 
^=i^[^-(3i^)2_  7r32]=1.227  sq.  inches,  while  fc2  for  a  nar- 
row circular  ring  like  the  present  section  may  be  put 
=  ^(1^)2  (see  §  98)  sq.  inches.  With  these  values,  and 
0=36,000  Ibs.  per  sq.  in.,  and  ^=3^5  (for  wrought  iron), 
we  have  from  eq.  (4),  for  flat  ends, 

P,=  _  1.227x36.000        =30743.6  lbg.    .     (1) 

i_i_  _  (144)- 

36,000  '  £  [1.625]' 


This  being  the  breaking  load,  the  safe  load  may  be  taken 
=  ^  or  yB  °f  30743.6  Ibs.,  according  as  the  structure  of 


376  MECHANICS   OF   ENGIKEERIXG. 

which  the  column  is  a  member  is  quiescent  or  subject  to 
vibration  from  moving  loads.  By  Hodgkinson's  formula 
33,768  Ibs.  was  obtained  as  a  breaking  load  in  this  case 
(§  306). 

For  rounded  ends  we  should  obtain  (eq.  5) 

P0=16,100.  Ibs.,  as  break  load  ^         (2) 

and  for  pin-and-square,  eq.  (6) 

P2= 24,908.  Ibs.  as  break,  load        .        .    (3) 

EXAMPLE  2.— (Same  as  Example  2,  §  306).     Eequired  by 

Rankine's  formula  the  necessary  diameter,  d,  to  be  given 

a  solid  cylindrical  cast-iron  pillar,  16  ft.  in  length,  with 

rounded  ends,  that  its  safe  load  may  be  six  long  tons  (i.e., 

of  2,240  Ibs.  each)  taking  6  as  a  factor  of  safety.     F=r~  , 

while  the  value  of  1&  is  thus  obtained.  From  §  247,  /  for 
a  full  circle  about  its  diameter  =%7:r4=7rr2.%rs  .:  Jc*=}£r3 
=y,6d2.  Hence  eq.  (5)  of  §  307  becomes. 


P0  the  breaking  load  is  to  be  =6x6x2,240  Ibs.,  0  for  cast- 
iron  is  80,000  Ibs.  per  sq.  inch,  while  ft  (abstract  number) 
— ejob  Solving  for  d  we  have  the  biquadratic  equation : 

d,_  28x6x6x2,240^  28x6x6x2,240xl62xl22x4 
22  x  80,000  22  x  80,000  x  400 

whence  c?2=  0.641  (1±  33.92),  and  taking  the  upper  sign, 
finally,  d=  A/22.4  =4.73  inches.  (By  Hodgkinson's  rule 
we  obtained  4.85  inches). 

309.  Radii  of  Gyration.  —  The  following  table,  taken  from 
p.  523  of  Rankine's  Civil  Engineering,  gives  values  of  &*, 
the  square  of  the  least  radius  of  gyration  of  the  given  cross- 
section  about  a  gravity-axis.  By  giving  the  least  value  of 


FLEXURE.   LONG  COLUMNS. 


377 


F  it  is  implied  that  the  plane  of  flexure  is  not  determined 
by  the  end-conditions  of  the  column ;  (i.e.,  it  is  implied 
that  the  column  has  either  flat  ends  or  round  ends.)  If 
either  end  (or  both)  is  a  pin-joint  the  column  may  need  to 
be  treated  as  having  a  flat-end  as  regards  flexure  in  a  plane 
containing  the  axis  of  the  column  and  the  axis  of  the  pin, 
if  the  bearings  of  the  pin  are  firm ;  while  as  regards  flex- 
ure in  a  plane  perpendicular  to  the  pin  it  is  to  be  consid- 
ered round-ended  at  that  extremity. 

In  the  case  of  a  "  thin  cell "  the  value  of  ^  is  strictly 
true  for  metal  infinitely  thin  and  of  uniform  thickness  ;  still, 
if  that  thickness  does  not  exceed  l/%  of  the  exterior  diame- 
ter, the  form  given  is  sufficiently  near  for  practical  pur- 
poses ;  similar  statements  apply  to  the  branching  forms. 


Solid  Eectangle. 

h=  least  side. 

Thin  Square  Cell 

Side=  h. 

Thin  Rectangular  Cell. 

A=  least  side. 

Solid  Circular  Section. 

Diameter  =d. 

Thin  Circular  Cell 

Exterior  diam.  =  d. 


Fig.  317  (a). 
Fig.  317(6). 
Fig.  317  (c). 
Fig.  317  (d). 
Fig.  317  (e). 


„_  A2     A+36 
~' 


Angle-Iron  of   Equal      Fig  3my) 
ribs 


378  MECHANICS   OF   ENGINEERING. 

Angle-Iron  of  unequal      Fig  o)>      jfc2= 


b-h2 


Fig.  318  (6).      #=-!* 


A 


Fig.  318  (c).      ^=^-3^ 


ribs. 

Cross  of  equal  arms. 
I-Beam  as  a  pillar. 
Let  area  of  web  =£. 
"        "     "  both  flanges 

=A. 

Channel 
Iron. 

Let  area  of  web  =B ;  of  flanges   =A  (both),     h  extends 
from  edge  of  flange  to  middle  of  web. 


Fig.  318  (d).      tf=h2  L-_ 


FIG.  319. 


310.  Built  Columns. — The  "  compression  members  "  of 
wrought-iron  bridge  trusses  are  generally  composed  of 
several  pieces  riveted  together,  the  most  common  forms 
being  the  Phoenix  column  (ring-shaped,  in  segments,)  and 
combinations  of  channels,  plates,  and  lattice,  some  of  which 
are  shown  in  Figs.  319  and  320. 

Experiments  on  full  size  columns  of  these  kinds  were 
made  by  the  U.  S.  Testing  Board  at  the  Watertown  Arse- 
nal about  1880. 

The  Phoenix  columns  ranged  from  8  in.  to  28  feet  in 
length,  and  from  1  to  42  in  the  value  of  the  ratio  of  length 
to  diameter.  The  breaking  loads  were  found  to  be  some- 
what in  excess  of  the  values  computed  from  Rankine's 
formula  ;  from  10  to  40  per  cent,  excess.  In  the  pocket- 
book  issued  by  the  Phoenix  company  they  give  the  follow- 
ing formula  for  their  columns,  (wrought-iron.) 


FLEXUKE.   LONG  COLUMNS.  379 


Breaking  load  in  Ibs.   )  _    50,000  F 
for  flat-ended  columns  )       1  _j_     j8 


3,OOOA2 

where  F  =  area  in  sq.  in.,  I  =  length,  and  A  =  external 
diameter,  both  in  the  same  unit. 

Many  different  formulae  have  been  proposed  by  different 
engineers  to  satisfy  these  and  other  recent  experiments  on 
columns,  but  all  are  of  the  general  form  of  Bankine's. 
For  instance  Mr.  Bouscaren,  of  the  Keystone  Bridge  Co., 
claims  that  the  strength  of  Phoenix  columns  is  best  given 
by  the  formula 

Breaking  load  in  )  _     38,000  F  fe)^ 

j  ~ 


Ibs.  for  flat-ends, 


1  + 


100,000/fc* 

(F  must  be  in  square  inches.) 

The  moments  of  inertia,  /,  and  thence  the  value  of  k*  = 
/  -f-  F,  for  such  sections  as  those  given  in  Figs.  319  and 
320  may  be  found  by  the  rules  of  §§  85-93,  (see  also  §  258.) 

311.  Moment  of  Inertia  of  Built  Column.  Example.  —  It  is  pro- 
posed to-  form  a  column  by  joining  two  I-beams  by  lattice- 
work, Fig.  321,  (a).  (While  the  lattice-work  is  relied  upon 
to  cause  the  beams  to  act  together  as  one  piece,  it  is  not 
regarded  in  estimating  the  area  F,  or  the  moment  of  iner- 
tia, of  the  cross  section).  It  is  also  required  to  find  the 
proper  distance  apart  =  x,  Fig.  321,  at  which  these  beams 
must  be  placed,  from  centre  to  centre  of  webs,  that  the 
liability  to  flexure  shall  be  equal  in  all  axial  planes,  i.e. 
that  the  1  of  the  compound  section  shall  be  the  same 
about  all  gravity  axes.  This  condition  will  be  ful- 
filled if  7Y  can  be  made  =  1^*  (§89),  0  being  the  centre 
of  gravity  of  the  compound  section,  and  X  perpendicular 
to  the  parallel  webs  of  the  two  equal  I-beams. 

Let  F'  =  the  sectional  area  of  one  of  the  I-beams,  Pv 
(see  Fig.  321(a)  its  moment  of  inertia  about  its  web-axis,  Jx' 
that  about  an  axis  "|  to  web.  (These  quantities  can  be 

*  That  is,  with  flat  ends  or  ball  ends  ;  but  with  pin  ends,  Fig.  312,  if  the 
pin  is  ||  to  X,  put  4/Y  =  /x  ;  if  II  to  T,  put  47X  =  /Y  . 


380 


MECHANICS   OF   ENGINEERING. 


found  in  the  hand-book  of  the  iron  company,  for  each  size 
of  rolled  beam). 
Then  the 

total  Jx  =  2/'x  ;  and  total  JY  =  2[>v  + 

(see  §88  eq.  4.)     If  these  are  to  be  equal,  we  write  them  so 
and  solve  for  x,  obtaining 


x  =     I'- 


(i) 


312.  Numerically;  suppose  each  girder  to  be  a  10^  inch 
light  I-beam,  105  Ibs.  per  yard,  of  the  N.  J.  Steel  and  Iron 
Co.,  in  whose  hand-book  we  find  that  for  this  beam  J'x  = 
185.6  biquad.  inches,  and  I'v  —  9.43  biquad.  inches,  while 
F'  =  10.44  sq.  inches.  "With  these  values  in  eq.  (1)  we 
have 


(185.6-9.43) 
10.44 


_  -v/67.5  =  8.21  inches. 


The  square  of  the  radius  of  gyration  will  be 

&2=2/'x-^2JF'=  371.2  -5-20.88  =17.7  sq.  in.       .       (2) 

and  is  the  same  for  any  gravity  axis  (see  §  89). 

As  an  additional  example,  suppose  the  two  I-beams  united 
by  plates  instead  of  lattice.  Let  the  thickness  of  the  plate 
=  t,  Fig.  321,  (b).  Neglect  the  rivet-holes.  The  distance 
a  is  known  from  the  hand-book.  The  student  may  derive 
a  formula  for  x,  imposing  the  condition  that  (total  7X)=  7T- 


FLEXURE.      LONG   COLUMNS.  381 

313.    Trussed  Girders. — When  a  horizontal  beam  is  trussed 


FIG.  323. 

in  the  manner  indicated  in  Fig.  322,  with  a  single  post  or 
strut  under  the  middle  and  two  tie-rods,  it  is  subjected  to 
a  longitudinal  compression  due  to  the  tension  of  the  tie- 
rods,  and  hence  to  a  certain  extent  resists  as  a  column,  the 
plane  of  whose  flexure  is  vertical,  (since  we  shall  here  sup- 
pose the  beam  supported  later ally. )Taking  the  case  of  uni- 
form loading,  (total  load  =  JP)and  supposing  the  tie-rods 
screwed  up  (by  sleeve  nuts)  until  the  top  of  the  post  is  on 
a  level  with  the  piers,  we  know  that  the  pressure  between 
the  post  and  the  beam  is  P'  '=  %  W  (see  §  273).  Hence 
by  the  parallelogram  of  forces  (see  Fig.  322)  the  tension 
in  each  tie-rod  is 


Q- 


_6        W 

16  *    cos  a 


(1) 


2  cos  a 
At  each  pier  the  horizontal  component  of  Q  is 

P=  Q  sin  a=—  JFtan  a 
16 

Hence  we  are  to  consider  the  half -beam  BO  as  a  "pin-and- 
square  "  column  under  a  compressive  force  P=5/is  W  tan  a} 
as  well  as  a  portion  of  a  continuous  girder  over  three 
equidistant  supports  at  the  same  level  and  bearing  a  uni- 
form load  W.  In  the  outer  fibre  of  the  dangerous  section, 
0,  (see  also  §  273  and  Fig.  278)  the  compression  per  sq. 
inch  due  to  both  these  straining  actions  must  not  exceed 
a  safe  limit,  E\  (see  §  251).  In  eq.  (6)  §  307,  where  P2  is 
tne  breaking  force  for  a  pin-and-square  column,  the  great- 


382  MECHANICS   OF   ENGINEERING. 

est  stress  in  any  outer  fibre  =  C  (  =  the  Modulus  of  Crush- 
ing) per  unit  of  area.  If  then  we  write  pwl.  instead  of  G 
in  that  equation,  and  6/16  JFtan  a  instead  of  P2  we  have 

j  max.  stress  due  )  _          _   5      JFtan  an«  ,    16  Q  Z2 
(  to  column  action  j  ""  PCO]~  Jg  •        ^  "  ~'  **  " 

while  from  eq.  (3),  p.  326,  we  have  (remembering  that  our 
present  W  represents  double  the  W  of  §  273). 

(  max.  stress  due  )  =       =  l_^e=  ^ 
{  to  girder  action  j      ^"[     16     7        16 

By  writing  pcol  .+_pKi=  R'=  a  safe  value  of  compression  per 
unit-area,  we  have  the  equation  for  safe  loading 


tan  «!+.   ^+-16JW    .    .    (2) 


Here  I  =  the  half  -span  OS,  Fig.  322,  e  =  the  distance  of 
outer  fibre  from  the  horizontal  gravity  axis  of  the  cross 
section,  k*  —  the  radius  of  gyration  of  the  section  referred 
to  the  same  axis,  while  F  =  area  of  section.  fj  should  be 
taken  from  the  end  of  §307. 

EXAMPLE.—  If  the  span  is  30  ft.  —  360  in.,  the  girder  a  15 
inch  heavy  I-beam  of  wrought  iron,  200  Ibs.  to  the  yard,  in 
which  e  =  }4  of  15  =  7l/2  inches,  ,F=20  sq.  in.,  and  F  = 
35.3  sq.  inches  (taken  from  the  Trenton  Co.'s  hand-book), 
required  the  safe  load  W>  the  strut  being  5  ft.  long. 
From  §307,  ft  =  1  :  36,000;  tan  «  =  15-J-5  =  3.00.  Hence, 
using  the  units  pound  and  inch  throughout,  and  putting 
R'  =  12,000  Ibs.  per  sq.  in.  =  max.  allowable  compression 
stress,  we  have  from  eq.  (2) 

16x20x12,000 

" 


16       1       JClflO 
"" 


O)h  .ISO 
.3  J"1"      3 


9  36,000  *  35.3    ""      35.3 

i.  e.,  69,111  Ibs.  besides  the  weight  of  the  beam. 

If  the  middle  support  had  been  a  solid  pier,  the  safe  load 
would  have  been  48  tons  ;  while  if  there  had  been  no 
middle  support  of  any  kind,  the  beam  would  bear  safely 


FLEXURE.   LONG  COLUMNS. 


383 


only  11.5  tons, 
the  strut)]. 


[Let  the  student  design  the  tie-rods  (and 


314.  Buckling  of  Web-Plates  in  Built  Girders. — In  §257  men- 
tion was  made  of  the  fact  that  very  high  web  plates  in 
built  beams,  such  as  /beams  and  box-girders,  might  need 
to  be  stiffened  by  riveting  T-irons  on  the  sides  of  the  web. 
(The  girders  here  spoken  of  are  horizontal  ones,  such  as 
might  be  used  for  carrying  a  railroad  over  a  short  span  of 
20  to  30  feet. 

An  approximate  method  of  determining  whether  such 
stiffening  is  needed  to  prevent  lateral  buckling  of  the  web, 
may  be  based  upon  Rankine's  formula  for  a  long  column 
and  will  now  be  given. 

In  Fig.  323  we  have,  free,  a  portion  of  a  bent  I-beam, 
between  two  vertical  sections  at  a  distance  apart=  hi  -= 
the  height  of  the  web.  In  such  a  beam  under  forces  L  to 
its  axis  it  has  been  proved  (§256)  that  we  may  consider 
the  web  to  sustain  all  the  shear,  J,  at  any  section,  and  the 
flanges  to  take  all  the  tension  and  compression,  which 
form  the  "stress-couple"  of  the  section.  These  couples 
and  the  two  shears  are  shown  in  Fig.  323,  for  the  two 
exposed  sections.  There  is  supposed  to  be  no  load  on  this 
portion  of  the  beam,  hence  the  shears  at  the  two  ends  are 


Ntl                1 

J 

I, 

1 

j 

h-          > 

£ 

+ 

equal.  Now  the  shear  acting  between  each  flange  and  the 
horizontal  edge  of  the  web  is  equal  in  intensity  per  square 
inch  to  that  in  the  vertical  edge  of  the  web ;  hence  if  the 
web  alone,  of  Fig.  323,  is  shown  as  a  free  body  in  Fig.  324, 
we  must  insert  two  horizontal  forces  =  «7,  in  opposite 


384  MECHANICS  OF  ENGINEERING. 

directions,  on  its  upper  and  lower  edges.  Each  of  these 
=  J  since  we  have  taken  a  horizontal  length  h^  =•  height 
of  web.  In  this  figure,  324,  we  notice  that  the  effect  of 
the  acting  forces  is  to  lengthen  the  diagonal  BD  and 
shorten  the  diagonal  AC,  both  of  those  diagonals  making 
an  angle  of  45°  with  the  horizontal. 

Let  us  now  consider  this  buckling  tendency  along  ACt 
by  treating  as  free  the  strip  A  C,  of  small  width  =  bt.  This 
is  shown  in  Fig.  325.  The  only  forces  acting  in  the  direc- 
tion of  its  length  AC  are  the  components  along  AC  of  the 
four  forces  J'  at  the  extremities.  AVe  may  therefore  treat 
the  strip  as  a  long  column  of  a  length  I  =  hi  V%  of  a  sec- 
tional area  F  =  bbl}  (where  b  is  the  thickness  of  the  web 
plate),  with  a  value  of  Ar2  =  y]2  62  (see  §  309),  and  with 
fixed  (or  flat)  ends.  Now  the  sum  of  the  longitudinal 
components  of  the  two  J'.'s  at  A  is  Q  =  2  J'  y2  \/2 

=  J'  V2;  but  J'  itself  =  ^.  b   */2  h  V2,  since  the  small 

rectangle  on  which  J'  acts  has  an  area  =  b  }4  \  V2,  and 
the  shearing  stress  on  it  has  an  intensity  of  (J  -f-  bh^  per 
unit  of  area.  Hence  the  longitudinal  force  at  each  end  of 
this  long  column  is 


(1) 


According  to  eq.  (4)  and  the  table  in  §  307,  the  safe  load 
(factor  of  safety  =  4)  for  a  wought-iron  column  of  this 
form,  with  flat  ends,  would  be  (pound  and  inch) 

P_  #Mi36,000       _  9,00066,  ,9. 

1    -  1  -  9JT  --  —  •    •    •       \pj 

11      +       *fli       11 

' 


If,   then,   in   any   particular    locality   of    the    girder  (of 
wrought-iron)  we  find  that  Q  is  >  Plt  i.e. 

if  L  is  >  -i'0005    ..  (pound  and  inch)    .    ,     (3) 


FLEXURE.   LONG  COLUMNS.  385 

then  vertical  stiffeners  will  be  required  laterally. 

When  these  are  required,  they  are  generally  placed  at 
intervals  equal  to  A1}  (the  depth  of  web),  along  that  part 
of  the  girder  where  Q  is  >  Plt 

EXAMPLE  Fig.  326.  —  Will  stiffening  pieces  be  required 
in  a  built  girder  of  20  feet  span,  bearing  a  uniform  load  of 
40  tons,  and  having  a  web  24  in.  deep  and  ^  in.  thick  ? 

From  §   242  we  know  that  the      _  w  -40  TONS 
greatest  shear,  ./max.,  is  close  to 
either  pier,  and  hence  we  investi-  |    [*-  —  10-- 
gate  that  part  of  the  girder  first. 

J    max.    =    y*    W  =   20    tons 
=40,000  Ibs. 
.-.  (inch  and  lb.),  see  (3), 

Pro  826. 


,  ......    (4) 

while,  see  (3),  (inch  and  pound), 


1  . 
^1,500  ' 

which  is  less  than  1666.66. 

Hence  stiffening  pieces  will  be  needed  near  the  extremi- 
ties of  the  girder.  Also,  since  the  shear  for  this  case  of 
loading  diminishes  uniformly  toward  zero  at  the  middle 
they  will  be  needed  from  each  end  up  to  a  distance  of 
-  of  10  ft.  from  the  middle. 


586  MECHANICS  OF   ENGINEERING. 


CHAPTER  VII. 


LINEAR  ARCHES  (OF  BLOCKWOKK). 


315.  A  Blockwork  Arch,  is  a  structure,  spanning  an  opening 
or  gap,  depending,  for  stability,  upon  the  resistance  to 
compresssion  of  its  blocks,  or  voussoirs,  the  material  of 
which,  such  as  stone  or  brick,  is  not  suitable  for  sustain- 
ing a  tensile  strain.  Above  the  voussoirs  is  usually 
placed  a  load  of  some  character,  (e.q.  a  roadway,)  whose 
pressure  upon  the  voussoirs  will  be  considered  as  vertical, 
only.  This  condition  is  not  fully  realized  in  practice, 
unless  the  load  is  of  cut  stone,  with  vertical  and  horizontal 
joints  resting  upon  voussoirs  of  corresponding  shape  (see 
Fig.  327),  but  sufficiently  so  to  warrant 
its  assumption  in  theory.  Symmetry 
of  form  about  a  vertical  axis  will  also 
be  assumed  in  the  following  treatment. 


316.  Linear  Arches. — For  purposes  of 
theoretical  discussion  the  voussoirs  of 
Fig.  327  may  be  considered  to  become 
PIQ.  327.  infinitely  small  and  infinite  in  number, 

thus  forming  a  "  linear  arch,"  while  retaining  the  same 
shapes,  their  depth  ~|  to  the  face  being  assumed  constant 
that  it  may  not  appear  in  the  formulae.  The  joints 
between  them  are  "1  to  the  curve  of  the  arch,  i.e.,  adjacent 
voussoirs  can  exert  pressure  on  each  other  only  in  the 
direction  of  the  tangent -line  to  that  curve. 


LINEAlt  ARCHES. 


387 


317.  Inverted  Catenary,  or  Linear  Arch  Sustaining  its  Own 
Weight  Alone. — Suppose  the  infinitely  small  voussoirs  to 
have  weight,  uniformly  distributed  along  the  curve,  weigh- 
ing q  Ibs.  per  running  linear  unit.  The  equilibrium  of 
such  a  structure,  Fig.  328,  is  of  course  unstable  but  theo- 
retically possible.  Required  the  form  of  the  curve  when 
equilibrium  exists.  The  conditions  of  equilibrium  are, 
obviously  :  1st.  The  thrust  or  mutual  pressure  T  between 
any  two  adjacent  voussoirs  at  any  point,  A,  of  the  curve 
must  be  tangent  to  the  curve  ;  and  2ndly,  considering  a 
portion  BA  as  a  free  body,  the  resultant  of  H0  the  pres- 


Fio.  328.  Fio.  329.  Fig.  330. 

sure  at  B  the  crown,  and  T  at  A,  must  balance  E  the  re- 
sultant of  the  II  vertical  forces  (i.e.,weights  of  the  elementary 
voussoirs)  acting  between  B  and  A. 

But  the  conditions  of  equilibrium  of  a  flexible,  inexten- 
sible  and  uniformly  loaded  cord  or  chain  are  the  very 
same  (weights  uniform  along  the  curve)  the  forces  being 
reversed  in.  direction.  Fig.  329.  Instead  of  compression 
we  have  tension,  while  the  II  vertical  forces  act  toward  in- 
stead of  away  from,  the  axis  X.  Hence  the  curve  of  equi- 
librium of  Fig.  328  is  an  inverted  catenary  (see  §  48)  whose 
equation  is 


(1) 


See  Fig.  330.  e  =  2.71828  the  Naperian  Base.  The  "par- 
ameter "  c  may  be  determined  by  putting  x  =  a,  the  half 
span,  and  y=  0  Y,  the  rise,  then  solving  for  c  by  successive 


388 


MECHANICS   OF   ENGINEERING. 


approximations.  The  " horizontal  thrust"  or  HQ,  is  =  gc, 
while  if  s  =  length  of  arch  OA,  along  the  curve,  the  thrust 
T  at  any  point  A  is 

From  the  foregoing  it  may  be  inferred  that  a  series  of  vous- 

soirs  of  finite  dimensions,   arranged 

so  as  to  contain  the  catenary  curve, 

with  joints  ~|  to  that  curve  and  of 

equal  weights  for  equal  lengths  of 

arc   will    be    in    equilibrium,    and 

moreover  in   stable   equilibrium  on 

account  of  friction,  and  the   finite  FIG.  331. 

width  of  the  joints  ;  see  Fig.  331. 

318.  Linear  Arches  under  Given  Loading. — The  linear  arches 
to  be  considered  further  will  be  treated  as  without  weight 
themselves  but  as  bearing  vertically  pressing  loads  (each 
voussoir  its  own). 

Problem. — Given  the  form  of  the  linear  arch,  itself,  it  is 
required  to  find  the  law  of  vertical  depth  of  loading  under 
which  the  given  linear  arch  will  be  in  equilibrium.  Fig. 
332,  given  the  curve  ABC,  i.e.,  the  linear  arch  itself,  re- 
quired the  form  of  the  curve  MON,  or  upper  limit  of  load- 
ing, such  that  the  linear  arch  ABC  shall  be  in  equilibrium 
under  the  loads  lying  between  the  two  curves.  The  load- 
ing is  supposed  homogeneous  and  of  constant  depth  ~|  to 
paper ;  so  that  the  ordinates  z  between  the  two  curves  are 
proportional  to  the  load  per  horizontal  linear  unit.  Assume 
a  height  of  load  z0  at  the  crown,  at  pleasure ;  then  required 
the  z  of  any  point  m  as  a  function  of  z0  and  the  curve 
ABC. 


LINEAR   ARCHES.  <389 

Practical  Solution. — Since  a  linear  arch  under  vertical 
pressures  is  nothing  more  than  the  inversion  of  the  curve 
assumed  by  a  cord  loaded  in  the  saine  way,  this  problem 
might  be  solved  mechanically  by  experimenting  with  a 
light  cord,  Fig.  333,  to  which  are  hung  other  tieavy  cords, 
or  bars  of  uniform  weight  per  unit  length,  and  at  equal 
horizontal  distances  apart  when  in  equilibrium.  By  varying 
the  lengths  of  the  bars,  and  their  points  of  attachment,  we 
may  finally  find  the  curve  sought,  MON.  (See  also  §  343.) 

Analytical  Solution. — Consider  the  structure  in  Fig.  334. 
A  number  of  rods  of  finite  length,  in  the  same  plane,  are  in 
equilibrium,  bearing  the  weights  P,  Plt  etc.,  at  the  con- 


FIG.  334. 


Fio.  335. 


necting  joints,  each  piece  exerting  a  thrust  T  against  the 
adjacent  joint.  The  joint  A,  (the  "  pin  "  of  the  hinge),  im- 
agined separated  from  the  contiguous  rods  and  hence  free, 
is  held  in  equilibrium  by  the  vertical  force  P  (a  load)  and 
the  two  thrusts  T  and  T',  making  angles  =  0  and  6'  with 
the  vertical ;  Fig.  335  shows  the  joint  A  free.  From  ^hor- 
izontal comps.)=0,  we  have. 

^sin  d—T'  sin  0'. 

That  is,  the  horizontal  component  of  the  thrust  in  any  rod 
is  the  same  for  all ;  call  it  H0.  .'. 


sin 


(1) 


390 


MECHANICS   OF   ENGINEERING. 


Now  draw  a  line  As  T  to  T'  and  write   S  ( compons.  I  to 
As)=0 ;  whence  P  sin  6'=  T  sin  £,  and  [see  (1)] 


(2) 


Let  the  rods  of  Fig.  334  become  infinitely  small  and  infi- 
nite in  number  and  the  load  continuous.  The  length  of 
each  rod  becomes  =ds  an  element  of  the  linear  arch,  ft  is 
the  angle  between  two  consecutive  ds's,  d  is  the  angle  be- 
tween the  tangent  line  and  the  vertical,  while  P  becomes 
the  load  resting  on  a  single  dx,  or  horizontal  distance  be- 
tween the  middles  of  the  two  cfe's.  That  is,  Fig.  336,  if 
7-=  weight  of  a  cubic  unit  of  the 
loading,  P=fzdx.  (The  lamina  of 
arch  and  load  considered  is  unity, 
"|  to  paper,  in  thickness.)  H0=a 
constant  =  thrust  at  crown  0  ; 
6=6',  and  sin  fi=ds+f>,  (since  the' 
angle  between  two  consecutive  tan- 
gents is  —  that  between  two  con- 
secutive radii  of  curvature).  Hence 
eq.  (2)  becomes 


but  dx=ds  sin 


Fio.  836. 


(3) 


Call  the  radius  of  curvature  at  the  crown  pQi  and  since 
there  Z=ZQ  and  0=90°,  (3)  gives  rVo— ^foJ  hence  (3)  may 
be  written 


z= 


p  sin3  0 


(4) 


This  is  the  law  of  vertical  depth  of  loading  required.  For 
a  point  of  the  linear  arch  where  the  tangent  line  is  verti- 
cal, sin  tf  =0  and  z  would  =  oo  ;  i.e.,  the  load  would  be  in- 


LINEAR  ARCHES. 


391 


finitely  high.     Hence,  in  practice,  a  full  semi-circle,  for  in- 
stance, could  not  be  used  as  a  linear  arch. 

319.  Circular  Arc  as  Linear  Arch. — As  an  example  of  the 
preceding  problem  let  us  ap- 
ply eq.  (4)  to  a  circular  arc, 
Fig.  337,  as  a  linear  arch. 
Since  for  a  circle  />  is  con- 
stant =  r,  eq.  (4)  reduces 


to 


(5) 


837.  Hence  the  depth  of  loading 

must  vary  inversely  as  the  cube  of  the  sine  of  the  angle  d 
made  by  the  tangent  line  (of  the  linear  arch)  with  the  ver- 
tical. 

To  find  the  depth  z  by  construction.— Having  z0  given,  G 
being  the  centre  of  the  arch,  prolong  Ca  and  mate  ab  = 
z0 ;  at  b  draw  a  ~|  to  Cb,  intersecting  the  vertical  through  a 
at  some  point  d ;  draw  the  horizontal  dc  to  meet  Ca  at 
some  point  c.  Again,  draw  ce  ~|  to  Cc,  meeting  ad  in  e ; 
then  ae  =  z  required  ;  a  being  any  point  of  the  linear  arch. 
For,  from  the  similar  right  triangles  involved,  we  have 

za=ab=ad,  sin  6=ac  sin  d.  sin  6=ae  sin  6  sin  d  sin  Q 


i.e.,  ae=z. 


Q.RD. 


[see  (5.)] 

320.  Parabola  as  Linear  Arch. — To  apply  eq.  4  §  318  to  a 
parabola  (axis  vertical)  as  linear  arch,  we  must  find  values 
of  p  and  p0  the  radii  of  curvature  at  any  point  and  the 
crown  respectively.  That  is,  in  the  general  formula, 


we  must  substitute  the  forms  for  the  first  and  second  dif- 
ferential co-efficients,  derived  from  the  equation  of  the 


392 


MECHANICS    OF    ENGINEERING. 


PIG.  338.  FIG.  339. 

curve  (parabola)  in  Fig.  338,  i.e.  from  x2  =  2  py ;  whence 
we  obtain 


-,or  cot  ft,—  —and — ^L— _ 
o  p        dx*      p 


Hence  p  =  p  cosec.»0,  i.e.  p= 

I+p         *  f 


(6) 


At  the  vertex  6  =  90°  .•.  p0  =  p.  Hence  by  substituting 
for  p  and  p0  in  eq.  (4)  of  §  318  we  obtain 

z =z0=  constant  [Fig.  339] (7) 

for  a  parabolic  linear  arch.  Therefore  the  depth  of  homo- 
geneous loading  must  be  the  same  at  all  points  as  at  the 
crown  ;  i.e.,  the  load  is  uniformly  distributed  with  respect 
to  the  horizontal.  This  result  might  have  been  antici- 
pated from  the  fact  that  a  cord  assumes  the  parabolic 
form  when  its  load  (as  approximately  true  for  suspension 
bridges)  is  uniformly  distributed  horizontally.  See  §  46 
in  Statics  and  Dynamics. 

321.  Linear  Arch  for  a  Given  Upper  Contour  of  Loading,  the 
arch  itself  being  the  unknown  lower  contour.  Given  the 
upper  curve  or  limit  of  load  and  the  depth  z0  at  crown,  re- 
quired the  form  of  linear  arch  which  will  be  in  equili- 
brium under  the  homogenous  load  between  itself  and  that 
upper  curve.  In  Fig.  340  let  MON  be  the  given  upper 
contour  of  load,  z0  is  given  or  assumed,^'  and  z"  are  the 
respective  ordinates  of  the  two  curves  BA C  and  MON. 
Required  the  eqation  of  BAG. 


LINEAR   ARCHES. 


393 


As  before,  the  loading  is  homogenous,  so  that  the 
weights  of  any  portions  of  it  are  proportional  to  the 
corresponding  areas  between  the  curves.  (Unity  thick- 
ness ~|  to  paper.)  Now,  Fig.  341,  regard  two  consecutive 
ds's  of  the  linear  arch  as  two  links  or  consecutive  blocks 
bearing  at  their  junction  m  the  load  dP  =  7-  (zr  +  z")  dx  in 
which  f  denotes  the  heaviness  of  weight  of  a  cubic  unit  of 
the  loading.  If  T  and  T  are  the  thrusts  exerted  on  these 
two  blocks  by  their  neighbors  (here  supposed  removed) 
we  have  the  three  forces  dP,  T  and  T',  forming  a  system 
in  equilibrium.  Hence  from  2X  =0. 


Tcos<p  =  T 


tind 


JF=0  gives  T  sin  <?>'—  T  sin  <p  =  dP 


(1) 


(2) 


From  (1)  it  appears  that  T  cos  <f  is  constant  at  all  points 
of  the  linear  arch  (just  as  we  found  in  §  318)  and  hence 
=  the  thrust  at  the  crown,  =  H,  whence  we  may  write 

T=H  -:-  cos  tp  and  T=H  +  cos  <p'    .    .    .    (3) 
Substituting  from  (3)  in  (2)  we  obtain 

H  (tan  <p'  —  tan  </>)=dP (4) 

But  tan  9'  =$^  and  tan  w'  =  dz'+^z>  ,  (dx  constant) 
dx  dx 

while  dP  =  f  (zr  -f- «")  dx.  Hence,  putting  for  convenience 
H  =•  ya2,  (where  a  =  side  of  an  imaginary  square  of  the 


394  MECHANICS  OF  ENGINEERING. 

loading,  whose  thickness  =  unity  and  whose  weight  =  H) 
we  have. 

§'=>'+*">    .......  <5> 

as  a  relation  holding  good  for  any  point  of  the  linear  arch 
which  is  to  be  in  equilibrium  under  the  load  included 
between  itself  and  the  given  curve  whose  ordinates  are  %", 
Fig.  340. 

322.  Example  of  Preceding.  Upper  Contour  a  Straight  Line.— 
Fig.  342.  Let  the  upper  contour  be  a  right  line  and  hor- 
izontal ;  then  the  z"  of  eq.  5  becomes  zero  at  all  points  of 
ON.  Hence  drop  the  accent  of  z'  in  eq.  (5)  and  we  have 


y?     a2 
Multiplying  which  by  dz  we  obtain 


This  being  true  of  the  z,  dz,  d?z  and  dx  of  each  element  of 
the  curve  O'B  whose  equation  is  desired,  conceive  it  writ- 
ten out  for  each  element  between  0'  and  any  point  m,  and 
put  the  sum  of  the  left-hand  members  of  these  equations 
=  to  that  of  the  right-hand  members,  remembering  that 
a2  and  dx2  are  the  same  for  each  element.  This  gives 


_   /•*_!    /• 
x1  I  a' 

c/  <te-0  «/  z— 


_  ..,  . 

dx1  I  a'  Oaf  2 


ct. 


acfz         _          \  «o  /     .    .    .    .    (7.) 


v  (-]- 


LINEAR   ARCHES. 
°  tt        N  o 


FIG.  342.  Fie.  843 

Integrating  (7.)  between  0'  and  any  point  m 


i.e.,  x=a  log..      *- 


•    •    (8)' 

•    (8.) 


or  z= 


(9.) 


This  curve  is  called  the  transformed  catenary  since  we  may 
obtain  it  from  a  common  catenary  by  altering  all  the  ordi- 
nates  of  the  latter  in  a  constant  ratio,  just  as  an  ellipse 
may  be  obtained  from  a  circle.  If  in  eq.  (9)  a  were  =  zfl 
the  curve  would  be  a  common  catenary. 

Supposing  20  and  the  co-ordinates  xl  and  z±  of  the  point 
B  (abutment)  given,  we  may  compute  a  from  eq.  8  by  put- 
ting x  =xl  and  z  =  «„  and  solving  for  a.  Then  the  crown- 
thrust  H  =  fa?  becomes  known,  and  a  can  be  used  in  eqs. 
(8)  or  (9)  to  plot  points  in  the  curve  or  linear  arch.  From 
eq.  (9)  we  have 

area    )        f  .      5?  f\  \    ,^,  1-JSf  Al^l  (10) 

OO'mn  \  -Vo«**=2  J0  [  ~          ~e 


,,     - 

dx+e  dx~  ^le~ 

Fig.  343. 

Call  this  area,  A.    As  for  the  thrusts  at  the  different 
joints  of  the  linear  arch,  see  Fig.  343,  we  have  crown- 

thrust  =H  =  r<#  .       .       ;      •  •       (ii) 

and  at  any  joint  m  the  thrust 


396  MECHANICS   OF   ENGINEERING. 

323.  Remarks. — The  foregoing  results  may  be  utilized 
with  arches  of  finite  dimensions  by  making  the  arch-ring 
contain  the  imaginary  linear  arch,  and  the  joints  T  to  the 
curve  of  the  same.  Questions  of  friction  and  the  resist- 
ance of  the  material  of  the  voussoirs  are  reserved  for  a 
succeeding  chapter,  (§  344)  in  which  will  be  advanced  a 
more  practical  theory  dealing  with  approximate  linear 
arches  or  "  equilibrium  polygons "  as  they  will  then  be 
called.  Still,  a  study  of  exact  linear  arches  is  valuable  on 
many  accounts.  By  inverting  the  linear  arches  so  far  pre- 
sented we  have  the  forms  assumed  by  flexible  and  inexten- 
sible  cords  loaded  in.  the  same  way- 


GRAPHICAL   STATICS.  39? 


CHAPTER    VIIL 


ELEMENTS    OF    GRAPHICAL,    STATICS. 


324.  Definition. — In  many  respects  graphical   processes 
have  duvantages  over  the  purely  analytical,  which  recom- 
mend their  use  in  many  problems  where  celerity  is  desired 
without  refined  accuracy.    One  of  these  advantages  is  that 
gross  errors  are  more  easily  detected,  and  another  that 
the  relations  of  the  forces,  distances,  etc.,  are  made  so 
apparent  to  the  eye,  in  the  drawing,  that  the  general  effect 
of  a  given  change  in  the  data  can  readily  be  predicted  at 
a  glance. 

Graphical  Statics  w  the  system  of  geometrical  construc- 
tions by  which  problems  in  Statics  may  be  solved  by 
the  use  of  drafting  instruments,  forces  as  well  as  distances 
being  represented  in  amount  and  direction  by  lines  on  the 
paper,  of  proper  length  and  position,  according  to  arbi- 
trary scales  ;  so  many  feat  of  distance  to  the  linear  inch  of 
paper,  for  example,  for  distances ;  and  so  many  pounds  or 
tons  to  the  linear  inch  of  paper  for  forces. 

Of  course  results  should  be  interpreted  by  the  same 
scale  as  that  used  for  the  data.  The  parallelogram  of 
forces  is  the  basis  of  all  constructions  for  combining  and 
resolving  forces. 

325.  Force  Polygons  and  Concurrent  Forces  in  a  Plane. — If  a 
material  point  is  in  equilibrium  under  three  forces  Pt  P2 
P3  (in  the  same  plane  of  course)  Fig.  344,  any  one  of  them, 


MECHANICS   OF   ENGINEERING. 


as  Pj,  must  be  equal  and  opposite  to  R  the  resultant  of 
the  other  two  (diagonal  of  their  parallelogram).  If  now 
we  lay  off  to  some  convenient  scale  a  line  in  Fig.  345  = 
Pj  and  ||  to  Pj  in  Fig.  344  ;  and  then  from  the  pointed  end 
of  PI  a  line  equal  and  ||  to  P2  and 

(laid  off  pointing  the  same  ivay,  we 
note   that   the  line   remaining   to 
i>  close  the  triangle  in  Fig.  345  must 
be  =  and  ||  to  P3,  since  that  tri- 
angle is   nothing   more   than   the 
left-hand     half-parallelogram     of 
Fm.344.  FIG.  345.  Fig.    344.      Also,    in  345,  *to  dose 

the  triangle  properly  the  directions  of  the  arrows  must 
be  continuous  Point  to  Butt,  round  the  periphery.  Fig. 
345  is  called  a  force  polygor  ;  of  three  sides  only  in  this 
case.  By  means  of  it,  given  any  two  of  the  three  forces 
which  hold  the  point  in  equilibrium,  the  third  can  be 
found,  being  equal  and  ||  to  the  side  necessary  to  "  close  " 
the  force  polygon. 

Similarly,  if  a  number  of  forces  in  a  plane  hold  a  mate- 
rial point  in  equilibrium,  Fig.  346,  their  force  polygon, 


FIG.  346. 


FIG.  347. 


Fig.  347,  must  close,  whatever  be  the  order  in  which  its 
sides  are  drawn.  For,  if  we  combine  Pt  and  P2  into  a  re- 
sultant Oa,  Fig.  346,  then  this  resultant  with  P3  to  form  a 
resultant  Ob,  and  so  on ;  we  find  the  resultant  of  Plt  P2,  P3, 
and  P4  to  be  Oc,  and  if  a  fifth  force  is  to  produce  equilib- 
rium it  must  be  equal  and  opposite  to  Oc,  and  would  close 
the  polygon  Odabc.O,  in  which  the  sides  are  equal  and  par- 


GRAPHICAL   STATICS.  399 

allel  respectively  to  the  forces  mentioned.  To  utilize  this 
fact  we  can  dispense  with  all  parts  of  the  parallelograms  in 
Fig.  346  except  the  sides  mentioned,  and  then  proceed  as 
follows  in  Fig.  347  : 

If  P5  is  the  unknown  force  which  is  to  balance  the  other 
four  (i.e,  is  their  anti-resultant],  we  draw  the  sides  of  the 
force  polygon  from  A  round  to  B,  making  each  line  paral- 
lel and  equal  to  the  proper  force  and  pointing  the  same 
way  ;  then  the  line  BA  represents  the  required  P5  in 
amount  and  direction,  since  the  arrow  BA  must  follow 
the  continuity  of  the  others  (point  to  butt). 

If  the  arrow  BA  were  pointed  at  the  extremity  B,  then 
it  gives,  obviously,  the  amount  and  direction  of  the  result- 
ant of  the  four  forces  Pl  .  .  .  P4.  The  foregoing  shows 
that  if  a  system  of  Concurrent  Forces  in  a  Plane  is  in  equi- 
librium, its  force  polygon  must  close. 

326.  Non-Concurrent  Forces  in  a  Plane. — Given  a  system  of 
non-concurrent  forces  in  a  plane,  acting  on  a  rigid  body, 
required  graphic  means  of  finding  their  resultant  and  anti- 
resultant  ;  also  of  expressing  conditions  of  equilibrium. 
The  resultant  must  be  found  in  amount  and  direction  ;  and 
also  in  position  (i.e.,  its  line  of  action  must  be  determined). 
E.  g.,  Fig.  348  shows  a  curved  rigid  beam  fixed  in  a  vise 
at  T,  and  also  under  the  action  of  forces  Pj  P2  P3  and  P4 
(besides  the  action  of  the  vise);  required  the  resultant  of 


By  the  ordinary 
parallelogram  of 
forces  we  com- 
bine Pl  and  P,  at 
a,  the  intersection 
of  their  lines  of 

FIG.  348.  action,  into  a  re- 

sultant .Z?a ;  then  7?a  with  P3  at  b,  to  form  Hb;  and  finally  7?b 
with  P4  at  c  to  form  Rc  which  is  .•.  the  resultant  required, 
i.e.,  of  Pl  .  .  .  .  Pt ;  and  c  .  .  .  F  is  its  line  of  action. 


MECHANICS  OF    ENGINEERING. 


The  separate  force  triangles  (half-parallelograms)  by 
which  the  successive  partial  resultants  R&,  etc.,  were  found, 
are  again  drawn  in  Fig.  349.  Now  since  Ec,  acting  in  the 

line c..F, Fig.  348, 
is  the  resultant  of 
PI  .  .  PI,  it  is  plain 
p<  that    a    force    Rc' 
equal  to  Rc  and  act- 
ing along  c .  .  Jf^but 
FIG-  349-  in  the  opposite  di- 

rection, would  balance  the  system  Pl . . .  P4,  (is  their  anti- 
resultant).  That  is,  the  forces  Pt  P2  P3  P4  and  Me'  would 
form  a  system  in  equilibrium.  The  force  fic'  then,  repre- 
sents the  action  of  the  vise  T  upon  the  beam.  Hence  re- 
place the  vise  by  the  force  Rc'  acting  in  the  line  .  .  .  F .  .  .  c  • 
to  do  which  requires  us  to  imagine  a  rigid  prolongation  of 
that  end  of  the  beam,  to  intersect  F . . .  c.  This  is  shown  in 
Fig.  350  where  the  whole  beam  is  free,  in  equilibrium,  under 
the  forces  shown,  and  in  precisely  the  same  state  of  stress, 
part  for  part,  as  in  Fig.  348.  Also,  by  combining  in  one 
force  diagram,  in  Fig.  351,  all  the  force  triangles  of  Fig.  349 
(by  making  their  common  sides  coincide,  and  putting  Rcf 
instead  of  7?c,  and  dotting  all  forces  other  than  those  of 
Fig.  350),  we  have  a  figure  to  be  interpreted  in  connection 
with  Fig.  350. 


o'   P, 


A|P2 


SPACE  DIAGRAM 
FIG.  350. 


FORCE  DIAGRAMS! 
FIG.  351. 


Here  we  note,  first,  that  in  the  figure  called  a  force-dia- 
gram, P!  P2  P3  JP4  and  R,.'  form  a  closed  polygon  and  that 


GRAPHICAL   STATICS.  401 

their  arrows  follow  a  continuous  order,  point  to  butt, 
around  the  perimeter  ;  which  proves  that  one  condition  of 
equilibrium  of  a  system  of  non-concurrent  forces  in  a  plane 
is  that  its  force  polygon  must  close.  Secondly,  note  that  ab 
is  ||  to  Oa',  and  be  to  Ob' ;  hence  if  the  force-diagram  has 
been  drawn  (including  the  rays,  dotted)  in  order  to  deter- 
mine the  amount  and  direction  of  Rc',  or  any  other  one  force, 
we  may  then  find  its  line  of  action  in  the  space-diagram,  as 
follows:  (N.  B. — By  space  diagram  is  meant  the  figure  show- 
ing to  a  true  scale  the  form  of  the  rigid  body  and  the  lines 
of  action  of  the  forces  concerned).  Through  a,  the  intersec- 
tion of  Pl  and  P2,  draw  a  line  ||  to  Oa'  to  cut  P3  in  some  point 
b  ;  then  through  b  a  line  ||  to  Ob'  to  cut  P4  at  some  point  c;  cF 
drawn  il  to  Oc'  is  the  required  line  of  action  of  7?c',  the  anti- 
resultant  of  P,,  P,,  P3,  and  P4. 

obc  is  called  an  equilibrium  polygon;  this  one  having  but 
two  segments,  ab  and  be  (sometimes  the  lines  of  action  of  Pl 
and  Rc'  may  conveniently  be  considered  as  segments.)  The 
segments  of  the  equilibrium  polygon  are  parallel  to  the  respect- 
ive rays  of  the  force  diagram. 

Hence  for  the  equilibrium  of  a  system  of  non-concurrent 
forces  in  a  plane  not  only  must  its  force  polygon  close, 
but  also  the  first  and  last  segments  of  the  corre- 
sponding equilibrium  polygon  must  coincide  with 
the  resultants  of  the  first  two  forces,  and  of  the  last 
two  forces,  respectively,  of  the  system.  E.g.,  ab  coin- 
cides with  the  line  of  action  of  the  resultant  of  Pl  and  P2 ; 
be  with  that  of  P4  and  R'c.  Evidently  the  equil.  polygon 
will  be  different  with  each  different  order  of  forces  in 
the  force  polygon  or  different  choice  of  a  pole,  0.  But  if 
the  order  of  forces  be  taken  as  above,  as  they  occur  along 
the  beam,  or  structure,  and  the  pole  taken  at  the  "  butt "  of 
the  first  force  in  the  force  polygon,  there  will  be  only  one  ; 
(and  this  one  will  be  called  the  special  equilibrium  polygon 
in  the  chapter  on  arch-ribs,  and  the  "  true  linear  arch  "  in 
dealing  with  the  stone  arch.)  After  the  rays  (dotted  in 
Fig.  351)  have  been  added,  by  joining  the  pole  to  each 


402 


MECHANICS    OF   ENGINEERING. 


vertex  with  which  it  is  not  already  connected,  the  final 
figure  may  be  called  the  force  diagram. 

It  may  sometimes  be  convenient  to  give  the  name  of 
rays  to  the  two  forces  of  the  force  polygon  which  meet 
at  the  pole,  in  which  case  the  first  and  last  segments  of 
the  corresponding  equil.  polygon  will  coincide  with  the 
lines  of  action  of  those  forces  in  the  space-diagram  (as  we 
may  call  the  representation  of  the  body  or  structure  on 
which  the  forces  act).  This  "  space  diagram  "  shows  the 
real  field  of  action  of  the  forces,  while  the  force  diagram, 
which  may  be  placed  in  any  convenient  position  on  the 
paper,  shows  the  magnitudes  and  directions  of  the  forces 
acting  in  the  former  diagram,  its  lines  being  interpreted 
on  a  scale  of  so  many  Ihs.  or  tons  to  the  inch  of  paper  ;  in 
the  space-diagram  we  deal  with  a  scale  of  so  many/ee£  to 
the  inch  of  paper. 

We  have  found,  then,  that  if  any  vertex  or  corner  of  the 
closed  force  polygon  be  taken  as  a  pole,  and  rays  drawn 
from  it  to  all  the  other  corners  of  the  polygon,  and  a  cor- 
responding equil.  polygon  drawn  in  the  space  diagram,  the 
first  and  last  segments  of  the  latter  polygon  must  co-incide 
with  the  first  and  last  forces  according  to  the  order 
adopted  (or  with  the  resultants  of  the  first  two  and  last 
two,  if  more  convenient  to  classify  them  thus).  It  remains 
to  utilize  this  principle. 

327.  To  Find  the  Resultant  of  Several  Forces  in  a  Plane.  —  This 
might  be  done  as  in  §  326,  but  since  frequently  a  given  set 
of  forces  are  parallel,  or  nearly  so,  a  special  method  will 
now  be  given,  of  great  convenience  in  such  cases.  Fig.  352. 

Let  P!  P2  and 


forces  whose 
resultant  is  re- 
q™ed.  Let  us 
first  find  their 
an^i  -  resultant, 
or  force  which 
will  balance 


GRAPHICAL  STATICS.  403 

them.  This  anti-resultant  may  be  conceived  as  decom- 
posed into  two  components  P  and  Pr  one  of  which,  say  P, 
is  arbitrary  in  amount  and  position.  Assuming  P,  then, 
at  convenience,  in  the  space  diagram,  it  is  required  to  find 
P'.  The  five  forces  must  form  a  balanced  system ;  hence 
if  beginning  at  Olt  Fig.  353,  we  lay  off  a  line  O^A  =  P  by 
scale,  then  A\  -  and  II  to  P,,  and  so  on  (point  to  butt),  the 
line  BOi  necessary  to  close  the  force  polygon  is  =  P'  re- 
quired. Now  form  the  corresponding  equil.  polygon  in 
the  space  diagram  in  the  usual  way,  viz.:  through  a  the 
intersection  of  P  and  PJ  draw  db  ||  to  the  ray  Oi  .  .  .  1 
(which  connects  the  pole  Ol  with  the  point  of  the  last  force 
mentioned).  From  b,  where  ab  intersects  the  line  of  P^ 
draw  be,  II  to  the  ray  0{  .  .  2,  till  it  intersects  the  line  of  P3. 
A  line  me  drawn  through  c  and  ||  to  the  P'  of  the  force 
diagram  is  the  line  of  action  of  P'. 

Now  the  resultant  of  P  and  P'  is  the  anti-resultant  of 
P1}  P2  and  P3; .'.  d,  the  intersection  of  the  lines  of  P  and 
P',  is  a  point  in  the  line  of  action  of  the  anti-resultant  re- 
quired, while  its  direction  and  magnitude  are  given  by  the 
line  BA  in  the  force  diagram  ;  for  B A  forms  a  closed  poly- 
gon both  with  P!  P2  P3,  and  with  PP'.  Hence  a  line 
through  d  \\  to  BA,  viz.,  de,  is  the  line  of  action  of  the  anti- 
resultant  (and  hence  of  the  resultant)  of  P,,  P2,  P3. 

Since,  in  this  construction,  P  is  arbitrary,  we  may  first 
choose  Oi,  arbitrarily,  in  a  convenient  position,  i.e.,  in  such 
a  position  that  by  inspection  the  segments  of  the  result- 
ing equil.  polygon  shall  give  fair  intersections  and  not 
pass  off  the  paper.  If  the  given  forces  are  parallel  the 
device  of  introducing  the  oblique  P  and  P'  is  quite  neces- 


328.— The  result  of  this  construction  may  be  stated  as 
follows,  (regarding  Oa  and  cm  as  segments  of  the  equil. 
polygon  as  well  as  ab  and  be):  If  any  two  segments  of  an 
equil  polygon  be  prolonged,  their  intersection  is  a  point  in 
the  line  of  action  of  the  resultant  of  those  forces  acting  at 


404 


MECHANICS  OF  ENGINEERING. 


the  vertices  intervening  between  the  given  segments.   Here, 
the  resultant  of  PI  P2  P3  acts  through  d. 

329.  Vertical  Reaction  of  Piers,  etc. — Fig.  354.  Given  the 
vertical  forces  or  loads  PI  P2  and  P3  acting  on  a  rigid  body 
(beam,  or  truss)  which  is  supported  by  two  piers  having 
smooth  horizontal  surfaces  (so  that  the  reactions  must  be 
vertical),  required  the  reactions  V0  and  Fn  of  the  piers. 
For  an  instant  suppose  V0  and  Vn  known ;  they  are  in 


FIG.  354. 


equil.  with  Pl  P2  and  P3.  The  introduction  of  the  equal 
and  opposite  forces  P  and  P'  in  the  same  line  will  not  dis- 
turb the  equilibrium.  Taking  the  seven  forces  in  the 
order  P  V0  Pl  P2  P3  Vn  and  P',  a  force  polygon  formed  with 
them  will  close  (see  (b)  in  Fig.  where  the  forces  which 
really  lie  on  the  same  line  are  slightly  separated).  With 
0,  the  butt  of  P,  as  a  pole,  draw  the  rays  of  the  force  dia- 
gram OA,  OB,  etc.  The  corresponding  equil."  polygon 
begins  at  a,  the  intersection  of  P  and  VQ  in  (a)  (the  space 
diagram),  and  ends  at  n  the  intersection  of  P'  and  Fn. 
Join  an.  Now  since  P  and  P'  act  in  the  same  line,  an 
must  be  that  line  and  must  be  ||  to  P  and  P'  of  the  force 
diagram.  Since  the  amount  and  direction  of  P  and  P'  are 
arbitrary,  the  position  of  the  pole  0  is  arbitrary,  while 
PI,  P2,  and  P3  are  the  only  forces  known  in  advance  in  the 
force  diagram. 

Hence  F0  and'F"n  may  be  determined  as  follows:  Lay  off 
the  given  loads  P15  P2,  etc.,  in  the  order  of  their  occur- 
rence in  the  space  diagram,  to  form  a  "  load-line  "  AD 


GRAPHICAL  STATICS 


405 


(see  (6.)  Fig.  354)  as  a  beginning  for  a  force-diagram  ;  take 
any  convenient  pole  0,  draw  the  rays  OA,  OB,  OC  and 
OD.  Then  beginning  at  any  convenient  point  a  in  the 
vertical  line  containing  the  unknown  V0,  draw  ab  ||  to  OA, 
be  ||  to  OB,  and  so  on,  until  the  last  segment  (dn  in  this 
case)  cuts  the  vertical  containing  the  unknown  Fn  in  some 
point  n.  Join  an  (this  is  sometimes  called  a  closing  line) 
and  draw  a  ||  to  it  through  0,  in  the  force-diagram.  This 
last  line  will  cut  the  "  load-line  "  in  some  point  n',  and 
divide  it  in  two  parts  n'  A  and  Dn',  which  are  respectively 
V0  and  Vu  required. 

Corollary. — Evidently,  for  a  given  system  of  loads,  in  given 
vertical  lines  of  action,  and  for  two  given  piers,  or  abut- 
ments, having  smooth  horizontal  surfaces,  the  location  of  the 
point  n'  on  the  load  line  is  independent  of  the  choice  of  a 
pole. 

Of  course,  in  treating  the  stresses  and  deflection  of  the 
rigid  body  concerned,  P  and  P'  are  left  out  of  account,  as 
being  imaginary  and  serving  only  a  temporary  purpose. 

330.  Application  of  Foregoing  Principles  to  a  Roof  Truss.- 
Eig.  355.  Wl  and  JF2  are  wind  pressures,  Pl  and  P2  aro 
loads,  while  the  remaining  external  forces,  viz.,  the  re- 


FIG.  355. 


40G  MECHANICS  OF   ENGINEERING. 

actions,  or  supporting  forces,  Vot  Va  and  Hn,  may  be  found 
by  preceding  §§.  (We  here  suppose  that  the  right  abut- 
ment furnishes  all  the  horizontal  resistance ;  none  at  the 
left). 

Lay  off  the  forces  (known)  Wlt  W2,  P1}  and  P.2  in  the 
usual  way,  to  form  a  portion  of  the  closed  force  polygon. 
To  close  the  polygon  it  is  evident  we  need  only  draw  a 
horizontal  through  5  and  limit  it  by  a  vertical  through  1. 
This  determines  HIL  but  it  remains  to  determine  n'  the 
point  of  division  between  V0  and  Va.  Select  a  convenient 
pole  0,,  and  draw  rays  from  it  to  1,  2,  etc.  Assume  a  con- 
venient point  a  in  the  line  of  Vn  in  the  space  diagram,  and 
through  it  draw  a  line  ||  to  Ojl  to  meet  the  line  of  Wl  in 
some  point  b  ;  then  a  line  II  to  0t2  to  meet  the  line  of  W2 
in  some  point  c ;  then  through  c  \\  to  0$  to  meet  the  line 
of  PI  in  some  point  d  ;  then  through  d  \\  to  0^  to  meet  the 
line  of  P2  in  some  point  e,  (e  is  identical  with  d,  since  P{ 
and  P2  are  in  the  same  line) ;  then  ef  \\  to  Ofi  to  meet  HD 
in  some  point/;  then/gr  ||  to  0,6  to  meet  Vn  in  some 
point  g. 

abcdefg  is  an  equilibrium  polygon  corresponding  to  the 
pole  Oi. 

Now  join  ag,  the  "  closing-line,"  and  draw  a  II  to  it 
through  Ol  to  determine  %',  the  required  point  of  division 
between  V0  and  Vn  on  the  vertical  1  6.  Hence  V0  and  Vn 
are  now  determined  as  well  as  Hn. 

[The  use  of  the  arbitrary  pole  0^  implies  the  temporary 
employment  of  a  pair  of  opposite  and  equal  forces  in  the  line 
ag,  the  amount  of  either  being  =  Ojw']. 

Having  now  all  the  external  forces  acting  *on  the  truss, 
and  assuming  that  it  contains  no  "  redundant  parts,"  i.e., 
parts  unnecessary  for  rigidity  of  the  frame-work,  we  proceed 
tc  find  the  pulls  and  thrusts  in  the  individual  pieces,  on 
the  following  plan.  The  truss  being  pin-connected,  no 
piece  extending  beyond  a  joint,  and  all  loads  being  con- 
sidered to  act  at  joints,  the  action,  pull  or  thrust,  of  each 
piece  on  the  joint  at  either  extremity  will  be  in  the  direction 
of  the  piece,  i.e.,  in  a  known  direction,  and  the  pin  of  each 


GRAPHICAL   STATICS.  407 

joint  is  in  equilibrium  under  a  system  of  concurrent  forces 
consisting  of  the  loads  (if  any)  at  the  joint  and  the  pulls 
or  thrusts  exerted  upon  it  by  the  pieces  meeting  there. 
Hence  we  may  apply  the  principles  of  §  325  to  each  joint 
in  turn.  See  Fig.  356.  In  constructing  and  interpreting 
the  various  force  polygons,  Mr.  R.  H  Bow's  convenient 
notation  will  be  used ;  this  is  as  follows  :  In  the  space 
diagram  a  capital  letter  [ABC,  etc.]  is  placed  in  each  tri- 
angular cell  of  the  truss,  and  also  in  each  angular  space  in 
the  outside  outline  of  the  truss  between  the  external  forces 
and  the  adjacent  truss-pieces.  In  this  way  we  can  speak  of 
the  force  Wl  as  the  force  BC,  of  W2  as  the  force  CE,  the 
stress  in  the  piece  a/3  as  the  force  CD,  and  so  on.  That 
is,  the  stress  in  any  one  piece  can  be  named  from  the 
letters  in  the  spaces  bordering  its  two  sides.  Corresponding 
to  these  capital  letters  in  the  spaces  of  the  space-dia- 
gram, small  letters  will  be  used  at  the  vertices  of  the  closed 
force-polygons  (one  polygon  for  each  joint)  in  such  a  way 
that  the  stress  in  the  piece  CD,  for  example,  shall  be  the 
force  cd  of  the  force  polygon  belonging  to  any  joint  in 
which  that  piece  terminates ;  the  stress  in  the  piece  FO 
by  the  force  fg  in  the  proper  force  polygon,  and  so  on. 

In  Fig.  356  the  whole  truss  is  shown  free,  in  equili- 
brium under  the  external  forces.  To  find  the  pulls  or 
thrusts  (i.e.,  tensions  or  compressions)  in  the  pieces,  con- 
sider that  if  all  but  two  of  the  forces  of  a  closed  force 
polygon  are  known  in  magnitude  and  direction,  while  the 
directions,  only,  of  those  two  are  known,  the  ichde  force 
polygon  may  be  drawn,  thus  determining  the  amounts  of 
those  two  forces  by  the  lengths  of  the  corresponding 
sides. 

We  must .-.  begin  with  a  joint  where  no  more  than  two 
pieces  meet,  as  at  a ;  [call  the  joints  a,  /9,  f,  d,  and  the  cor- 
correspouding  force  polygons  a',  $'  etc.  Fig.  356.]  Hence 
at  «'  (anywhere  on  the  paper)  make  ab  \\  and  =  (by  scale) 
to  the  known  force  AB  (i.e.,  F0)  pointing  it  at  the  upper  end, 
and  from  this  end  draw  be  =  and  ||  to  the  known  force  BC 
(i.e.,  W,)  pointing  this  at  the  lower  end. 


408 


MECHANICS   OP   ENGINEERING. 


FIG    356. 

To  close  the  polygon  draw  through  c  a' II  to  the  piece 
CD,  and  through  a  a  ||  to  AD ;  their  intersection  deter- 
mines d,  and  the  polygon  is  closed.  Since  the  arrows 
must  be  point  to  butt  round  the  periphery,  the  force  with 
which  the  piece  CD  acts  on  the  pin  of  the  joint  a  is  a 
force  of  an  amount  =  cd  and  in  a  direction  from  c  toward 
d ;  hence  the  piece  CD  is  in  compression  ;  whereas  the 
action  of  the  piece  DA  upon  the  pin  at  a  is  from  d  toward 
a  (direction  of  arrow)  and  hence  DA  is  in  tension.  Notice 
that  in  constructing  the  force  polygon  a'  a  right-handed 
(or  clock-wise)  rotation  has  been  observed  in  considering 
in  turn  the  spaces  AEG  and  D,  round  the  joint  a.  A 
similar  order  will  be  found  convenient  in  each  of  the  other 
joints. 

Knowing  now  the  stress  in  the  piece  CD,  (as  well  as  in 
DA)  all  but  two  of  the  forces  acting  on  the  pin  at  the  joint 
ft  are  known,  and  accordingly  we  begin  a  force  polygon,  /j', 
for  that  joint  by  drawing  dc,=  and  ||  to  the  dc  of  polygon 
a',  but  pointed  in  the  opposite  direction,  since  the  action  of 
CD  on  the  joint  ft  is  equal  and  opposite  to  its  action  on 
the  joint  «  (this  disregards  the  weight  of  the  piece). 
Through  c  draw  ce  —  and  ||  to  the  force  CE  (i.e.,  W2)  and 


GRAPHICAL   STATICS.  409 

pointing  the  same  way ;  then  ef,  =  and  ||  to  the  load  EF 
(i.e.  PI)  and  pointing  downward.  Through  f  draw  a  ||  to 
the  piece  FG  and  through  d,  a  ||  to  the  piece  GD,  and  the 
polygon  is  closed,  thus  determining  the  stresses  in  the 
pieces  FG  and  GD.  Noting  the  pointing  of  the  arrows, 
we  readily  see  that  FG  is  in  compression  while  GD  is  in 
tension. 

Next  pass  to  the  joint  d,  and  construct  the  polygon  d', 
thus  determining  the  stress  gh  in  GH  and  that  ad  in  AD  ; 
this  last  force  ad  should  check  with  its  equal  and  oppo- 
site ad  already  determined  in  polygon  a'.  Another  check 
consists  in  the  proper  closing  of  the  polygon  f,  all  of 
whose  sides  are  now  known. 

[A  compound  stress-diagram  may  be  formed  by  super- 
posing the  polygons  already  found  in  such  a  way  as  to 
make  equal  sides  co-incide ;  but  the  character  of  each 
stress  is  not  so  readily  perceived  then  as  when  they  are 
kept  separate]. 

In  a  similar  manner  we  may  find  the  stresses  in  any  pin- 
connected  frame-work  (in  one  plane  and  having  no  redun- 
dant pieces)  under  given  loads,  provided  all  the  support- 
ing forces  or  reactions  can  be  found.  In  the  case  of  a 

braced-arch  (truss)  as 
shown  in  Fig.  357,  hinged 
to  the  abutments  at  both 
ends  and  not  free  to  slide 
laterally  upon  them,  the 
reactions  at  0  and  B  de- 
pend,  in  amount  and  direc- 
tion, not  only  upon  the-  equations  of  Statics,  but  on  the 
form  and  elasticity  of  the  arch-truss.  Such  cases  will  be 
treated  later  under  arch-ribs,  or  curved  beams. 

332.  The  Special  Equil.  Polygon.  Its  Relation  to  the  Stresses 
in  the  Rigid  Body. — Eeproducing  Figs.  350  and  351  in  Figs. 
358  and  359,  (where  a  rigid  curved  beam  is  in  equilibrium 
under  the  forces  P,,  P2,  Ps,  P4  and  R',.}  we  call  a  .  .  b  .  .  c 


410 


MECHANICS   OF   ENGINEERING. 


the  special  equil.  polygon  because  it  corresponds  to  a  force- 
diagram  in  which  the  same  order  of  forces  has  been  ob- 
served as  that  in  which  they  occur  along  the  beam  (from 
left  to  right  here).  From  the  relations  between  the  force 


SPACE  DIAGRAM 

FIG.  358. 


FORCE  DIAGRAM 
FIG.  3o9. 


diagram  and  equil.  polygon,  this  special  equil.  polygon  in 
the  space  diagram  has  the  following  properties  in  connec- 
tion with  the  corresponding  rays  (dotted  lines)  in  the  force 
diagram. 

The  stresses  in  any  cross-section  of  the  portion  O'A  of 
the  beam,  are  due  to  P±  alone ;  those  of  any  cross-section 
on  AB  to  P!  and  P2,  i.e.,  to  their  resultant  Ra,  whose  mag- 
^itude  is  given  by  the  line  Oa'  in  the  force  diagram,  while 
its  liue  of  action  is  ab  the  first  segment  of  the  equil.  poly- 
gon. Similarly,  the  stresses  in  BG  are  due  to  PI}  P2  and 
Ps,  i.e.,  to  their  resultant  7?b  acting  along  the  segment  be, 
its  magnitude  being  =  Ob'  in  the  force  diagram.  E.g.,  if 
the  section  at  ra  be  exposed,  considering  O'ABm  as  a  free 
body,  we  have  (see  Fig.  360)  the  elastic  stresses  for  inter- 


nal forces)  at  m  balancing  the  exterior  or  "  applied  forces  " 
PI,  P2  and  P3.     Obviously,  then,  the  stresses  at  m  are  just 


GRAPHICAL    STATICS.  411 

the  same  as  if  JRb  the  resultant  of  Plt  P2  and  P3,  acted  upon 
an  imaginary  rigid  prolongation  of  the  beam  intersecting 
be  (see  Fig.  361).7?b  might  be  called  the  "anti-stress-result- 
ant "  for  the  portion  BC  of  the  beam.  We  may  .•.  state 
the  following :  If  a  rigid  body  is  in  equilibrium  under  a  sys- 
tem of  Non-Concurrent  Forces  in  a  plane,  and  the  special  equi- 
librium polygon  has  been  drawn,,  then  each  ray  of  the  force 
diagram  is  the  anti-stress-resultant  of  that  portion  of  the  beam 
ichich  corresponds  to  the  segment  of  the  equilibrium  polygon 
to  i. cinch  the  ray  is  parallel ;  and  its  line  of  action  is  the  seg- 
ment just  mentioned. 

Evidently  if  the  body  is  not  one  rigid  piece,  but  com- 
posed of  a  ring  of  uncemented  blocks  (or  voussoirs),  it  may 
be  considered  rigid  only  so  long  as  no  slipping  takes  place 
or  disarrangement  of  the  blocks ;  and  this  requires  that  the 
"  anti-stress-resultant "  for  a  given  joint  between  two 
blocks  shall  not  lie  outside  the  bearing  surface  of  the 
joint,  nor  make  too  small  an  angle  with  it,  lest  tipping  or 
slipping  occur.  For  an  example  of  this  see  Fig.  362,  show- 
ing a  line  of  three  blocks  in  equilibrium  under  five  forces. 

The  pressure  borne  at  the 
joint  MN,  is  =  R&  in  the 
force-diagram  and  acts  in 
the  line  ab.  The  con- 
struction supposes  all 
the  forces  given  except 
FIG.  362.  one,  in  amount  and  posi- 

tion, and  that  this  one  could  easily  be  found  in  amount,  as 
being  the  side  remaining  to  close  the  force  polygon,  while 
its  position  would  depend  on  the  equil.  polygon.  But  in 
practice  the  two  forces  Pl  and  R',.  are  generally  unknown, 
hence  the  point  0,  or  pole  of  the  force  diagram,  can  not 
be  fixed,  nor  the  special  equil.  polygon  located,  until  other 
considerations,  outside  of  those  so  far  presented,  are 
brought  into  play.  In  the  progress  of  such  a  problem,  as 
will  be  seen,  it  will  be  necessary  to  use  arbitrary  trial  po- 
sitions for  the  pole  0,  and  corresponding  trial  equilibrium 
polygons. 


MECHANICS  OF  ENGINEElilifG. 


CHAPTER  IX. 


GRAPHICAL  STATICS  OF  VERTICAL  FORCES. 


333.  Remarks.— (With  the  exception  of  §  378  a)  in  prob- 
lems to  be  treated  subsequently  (either  the  stiff  arch -rib,, 
or  the  block-work  of  an  arch-ring,  of  masonry)  when  the 
body  is  considered  free  all  the  forces  holding  it  m  equil. 
will  be  vertical  (loads,  due  to  gravity)  except  the  reactions 
at  the  two  extremities,  as  in  Fig.  363 ;  but  for  convenience 
each  reaction  will  be  replaced  by  its  horizontal  and  verti- 
cal components  (see  Fig.  364).  The  two  TTs  are  of  course 
equal,  since  they  are  the  only  horizontal  forces  in  the 
system.  Henceforth,  all  equil.  polygons  under  discussion  will 
be  understood  to  imply  this  kind  of  system  of  forces.  Pl}  P.>, 


•\ 


etc.,  will  represent  the  "  loads  "  ;  V0  and  Vn  the  vertical 
components  of  the  abutment  reactions  ;  //  the  value  of 
either  horizontal  component  of  the  same.  (We  here  sup- 
pose the  pressures  T0  and  TD  resolved  along  the  horizon- 
tal and  vertical.) 


GRAPHICAL   STATICS. 


413 


334.  Concrete  Conception  of  an  Equilibrium  Polygon. — Any 
equilibrium  polygon  has  this  property,  due  to  its  mode 
of  construction,  viz.:  If  the  ab  and  be  of  Fig.  358  were  im- 
ponderable straight  rods,  jointed  at  b  without  friction,  they 
would  be  in  equilibrium  under  the  system  of  forces  there 
given.  (See  Fig.  364a).  The  rod  ab  suffers  a  compression 
equal  to  the  RA  of  the  force  diagram,  Fig.  359,  and  be  a 
compression  =  Rb.  In  some  cases  these  rods  might  be  in 
tension,  and  would  then  form  a  set  of  links  playing  the 
part  of  a  suspension-bridge  cable.  (See  §  44). 

335.  Example  of  Equilibrium  Polygon  Drawn  to  Vertical  Loads 
— Fig.  365.  [The  structure  bearing  the  given  loads  is  not 
shown,  but  simply  the  imaginary  rods,  or  segments  of  an 
equilibrium  polygon,  which  would  support  the  given  loads 
in  equilibrium  if  the  abutment  points  A  and  JB,  to  which 
the  terminal  rods  are  hinged,  were  firm.  In  the  present 
case  this  equilibrium  is  unstable  since  the  rods  form  a 
standing  structure ;  but  if  they  were  hanging,  the  equilibri- 
um would  be  stable.  Still,  in  the  present  case,  a  very  light 
bracing,  or  a  little  friction  at  all  joints  would  make  the 
equilibrium  stable. 


p,   .1  ; 


Given  three  loads  Plt  P2,  and  P3,  and  two  "  abutment 
verticals  "  A'  and  B',  in  which  we  desire  the  equil.  poly- 
gon to  terminate,  lay  off  as  a  "load-line,"  to  scale,  P,,  P2, 
and  P3  end  to  end  in  their  order.  Then  selecting  any  pole, 


414  MECHANICS  OF   ENGINEERING. 

O,  draw  the  rays  01,  02,  etc.,  of  a  force  diagram  (the  F's 
and  P's,  though  really  on  the  same  vertical,  are  separated 
slightly  for  distinctness  ;  also  the  ZTs,  which  both  pass 
through  0  and  divide  the  load-line  into  F0  and  Fn).  We 
determine  a  corresponding  equilibrium  polygon  by  draw- 
ing through  A  (any  point  in  A'}  a  line  ||  to  0  .  .  1,  to  inter- 
sect Pi  in  some  point  b ;  through  6  a  ||  to  0  .  .  2,  and  so  on* 
until  B'  the  other  abutment-vertical  is  struck  in  some 
point  B.  AB  is  the  "  abutment-line  "  or  "  closing-line" 

By  choosing  another  point  for  0,  another  equilibrium 
polygon  would  result.  As  to  which  of  the  infinite 
number  (which  could  thus  be  drawn,  for  the  given  loads 
and  the  A  and  B'  verticals)  is  the  special  equilibrium  poly- 
gon for  the  arch-rib  or  stone-arch,  or  other  structure,  on 
which  the  loads  rest,  is  to  be  considered  hereafter.  In 
any  of  the  above  equilibrium  polygons  the  imaginary 
series  of  jointed  rods  would  be  in  equilibrium. 

336.  Useful  Property  of  an  Equilibrium  Polygon  for  Vertical 
Loads.— (Particular  case  of  §  328).  See  Fig.  366.  In  any 
equil.  polygon,  supporting  vertical  loads,  consider  as  free 
any  number  of  consecutive  segments,  or  rods,  with  the 
loads  at  their  joints,  e.  g.,  the  5th  and  6th  and  portions  of 
the  4th  and  7th  which,  we  sup- 
pose cut  and  the  compressive 
forces  in  them  put  in,  T\  and 
in  order  to  consider  4567 
a  free  body.  For  equil., 
according  to  Statics,  the  lines 
of  action  of  rl\  and  T7  (the  com- 
pression in  those  rods)  must  in- 
tersect in  a  point,  G,  in  the  line  of  action  of  the  resultant 
of  P4,  P5,  and  P6  ;  i.e.,  of  the  loads  occurring  at  the -inter- 
vening vertices.  That  is,  the  point  C  must  lie  in  the  ver- 
tical containing  the  centre  of  gravity  of  those  loads.  Since 
the  position  of  this  vertical  must  be  independent  of  the 
particular  equilibrium  polygon  used,  any  other  (dotted 
lines  in  Fig.  366)  for  the  same  loads  will  give  the  same  re- 


GRAPHICAL   STATICS.  415 

suits.  Hence  the  vertical  CD,  containing  the  centre  of 
gravity  of  any  number  of  consecutive  loads,  is  easily  found 
by  drawing  the  equilibrium  polygon  corresponding  to 
any  convenient  force  diagram  having  the  proper  load-line. 
This  principle  can  be  advantageously  applied  to  finding 
a  gravity -line  of  any  plane  figure,  by  dividing  the  latter 
into  parallel  strips,  whose  areas  may  be  treated  as  loads 
applied  in  their  respective  centres  of  gravity.  If  the  strips 
are  quite  numerous,  the  centre  of  gravity  of  each  may  be 
considered  to  be  at  the  centre  of  the  line  joining  the  mid- 
dles of  the  two  long  sides,  while  their  areas  may  be  taken 
as  proportional  to  the  lengths  of  the  lines  drawn  through 
these  centres  of  gravity  parallel  to  the  long  sides  and  lim- 
ited by  the  end-curves  of  the  strips.  Hence  the  "  load- 
line  "  of  the  force  diagram  may  consist  of  these  lines,  or 
of  their  halves,  or  quarters,  etc.,  if  more  convenient  (§  376). 


USEFUL  RELATIONS  BETWEEN  FORCE  DIA- 
GRAMS AND  EQUILIBRIUM  POLYGONS, 
(for  vertical  loads.) 

337.  R6sum6  of  Construction.— Fig.  367.  Given  the  loads 
PI,  etc.,  their  verticals,  and  the  two  abutment  verticals  A' 
and  B',  in  which  the  abutments  are  to  lie ;  we  lay  off  a 
load-line  1  ...  4,  take  any  convenient  pole,  0,  for  a  force- 
diagram  and  complete  the  latter.  For  a  corresponding 
equilibrium  polygon,  assume  any  point  A  in  the  vertical 
A',  for  an  abutment,  and  draw  the  successive  segments 
Al,  2,  etc.,  respectively  parallel  to  the  inclined  lines  of  the 
force  diagram  (rays),  thus  determining  finally  the  abut- 
ment JS,  in  B",  which  (B)  will  not  in  general  lie  in  the  hor- 
izontal through  A. 

Now  join  AB,  calling  AB  the  abutment-line,  and  draw  a 
parallel  to  it  through  0,  thus  fixing  the  point  n'  on  the 


416 


MECHANICS   OF    ENGINEERING. 


g 


i 

FIG. 


load-line.  This  point  n',  as  above  determined,  is  indepen- 
dent of  the  location  of  the  pole,  0,  (proved  in  §  329)  and 
divides  the  load-line  into  two  portions  (  V0  =  1 .  .  .  n',  and 
V'n  =  n'  ...  4)  which  are  the  vertical  pressures  which  two 
supports  in  the  verticals  A'  and  B'  ivould  sustain  if  the 
given  loads  rested  on  a  horizontal  rigid  bar,  as  in  Fig.  368. 

See  §  329.  Hence  to  find  the  point  n'  we  may  use  any 
convenient  pole  0. 

[N.  B.— The  forces  Vtt  and  FD  of  Fig.  367  are  not  identi- 
cal with  V0  and  F'n,  but  may  be  obtained  by  dropping  a 
"]  from  0  to  the  load-line,  thus  dividing  the  load-line 
into  two  portions  which  are  V '„  (upper  portion)  and  Vn. 
However,  if  A  and  B  be  connected  by  a  tie-rod,  in  Fig. 
367,  the  abutments  in  that  figure  will  bear  vertical  press- 
ures only  and  they  will  be  the  same  as  in  Fig.  368,  while 
the  tension  in  the  tie -rod  will  be  =  On'.] 

338.  Theorem. —  The  vertical  dimensions  of  any  two  'equili- 
brium polygons,  drawn  to  the  same  loads,  load-verticals,  and 
abutment-verticals,  are  inversely  proportional  to  their  H's  (or 
"  pole  distances  ").  We  here  regard  an  equil.  polygon  and 
its  a,butment-line  as  a  closed  figure.  Thus,  in  Fig.  369, 
we  have  two  force-diagrams  (with  a  common  load-line,  for 
convenience)  and  their  corresponding  equil.  polygons,  for 
the  same  loads  and  verticals.  From  §  337  we  know  that 
On'  is  ||  to  AB  and  00n'  is  ||  to  A0S0.  Let  CD  be  any  ver- 
tical cutting  the  first  segments  of  the  two  equil.  polygons. 


GKAPHICAL   STATICS. 


417 


Denote  the  intercepts  thus  determined  by  2'  and  «'0,  respect- 
ively. From  the 
parallelisms  just 
mentioned,  and 
others  more  famil- 
iar, we  have  the 
triangle  0  \n'  sim- 
ilar to  the  triangle 
Az'  (shaded),  and 
the  triangle  00ln' 
similar  to  the  tri- 
angle A0z£.  Hence 


FIG.  369. 


^0) 
~h   f 


the  proportions  between  (  In'  _zf       -,  lw' 
bases  and  altitudes  (  ~ff      ^          ~Jfo 

.-.  zf  :  z'0  :  :  H0  :  H.          The  same  kind  of  proof  may  easily 
be  applied  to  the  vertical  intercepts  in  any  other  segments, 


e.  g.,  z"  and  z" 


Q.  E.  D. 


339.    Corollaries  to  the  foregoing.     It  is  evident  that  : 
(1.)     If  the  pole  of  the  force-diagram  be  moved  along  a 
vertical  line,  the  equilibrium  polygon  changing  its  form 
in  a  corresponding  manner,  the  vertical  dimensions  of  the 
equilibrium  polygon  remain  unchanged  ;  and 

(2.)  If  the  pole  move  along  a  straight  line  which  con- 
tains the  point  n',  the  direction  of  the  abutment-line 
remains  constantly  parallel  to  the  former  line,  while  the 
vertical  dimensions  of  the  equilibrium  polygon  change  in 
inverse  proportion  to  the  pole  distance,  or  H,  of  the  force- 
diagram.  [His  the  ~|  distance  of  the  pole  from  the  load- 
line,  and  is  called  the  pole-distance]. 

§  340.  Linear  Arch  as  Equilibrium  Polygon.—  (  See  §  316.) 
If  the  given  loads  are  infinitely  small  with  infinitely  small 
horizontal  spaces  between  them,  any  equilibrium  polygon 
becomes  a  linear  arch.  Graphically  we  can  not  deal  with 
these  infinitely  small  loads  and  spaces,  but  from  §  336  it 
is  evident  that  if  we  replace  them,  in  successive  groups, 


418 


MECHANICS  OF   ENGINEERING. 


by  finite  forces,  each  of  which  =  the  sum  of  those  com- 
posing one  group  and  is 
applied  through  the  cen- 
of    gravity    of    that 


tre  of  gravity  of 
group,  we  can  draw  an 
equilibrium  polygon 
whose  segments  will  be 
tangent  to  the  curve  of 
the  corresponding  linear 
arch,  and  indicate  its  posi- 
tion with  sufficient  exactness  for  practical  purposes.  (See 
Fig.  370).  The  successive  points  of  tangency  A,  m,  n,  etc., 
lie  vertically  under  the  points  of  division  between  the 
groups.  This  relation  forms  the  basis  of  the  graphical 
treatment  of  voussoir,  or  blockwork,  arches. 

341.  To  Pass  an  Equilibrium  Polygon  Through  Three  Arbitrary 
Points. — (In  the  present  case  the  forces  are  vertical.  For 
a  construction  dealing  with  any  plane  system  of  forces  see 
construction  in  §  3780.)  Given  a  system  of  loads,  it  is  re- 
quired to  draw 

r/1 

-.2 


an  equilibrium 
polygon  for 
them  through 
any  three  points, 
two  of  which 
may  be  consid- 
ered as  abut- 
ments, outside  of  the  load- verticals,  the  third  point  being 
between  the  verticals  of  the  first  two.  See  Fig.  371.  The 
loads  Plt  etc.,  are  given,  with  their  verticals,  while  A,  p, 
and  B  are  the  three  points.  Lay  off  the  load-line,  and 
with  any  convenient  pole,  01?  construct  a  force-diagram, 
then  a  corresponding  preliminary  equilibrium  polygon 
beginning  at  A.  Its  right  abutment  Blt  in  the  vertical 
through  B,  is  thus  found.  Ov  n'  can  now  be  drawn  ||  to  ABV 
to  determine  n'.  Draw  n'O  II  to  BA.  The  pole  of  the 
required  equilibrium  polygon  must  lie  on  n'O  (§  337)- 


FIG.  371. 


GRAPHICAL   STATICS. 


419 


Draw,  a  vertical  through  p.  The  H  of  the  required  equili- 
brium polygon  must  satisfy  the  proportion  H :  H^  :  :  ~rs  : 
pin.  (See  §  338).  Hence  construct  or  compute  H  from 
the  proportion  and  draw  a  vertical  at  distance  H  from 
the  load-line  (on  the  left  of  the  load-line  here) ;  its  inter- 
section with  n'  0  gives  0  the  desired  pole,  for  which  a 
force  diagram  may  now  be  drawn.  The  corresponding 
equilibrium  polygon  beginning  at  the  first  point  A  will 
also  pass  through  p  and  B ;  it  is  not  drawn  in  the  figure. 

342.  Symmetrical  Case  of  the  Foregoing  Problem.— If  two 
points  A  and  B  are  on  a  level,  the  third,  p,  on  the  middle 
vertical  between  them  ;  and  the  loads  (an  even  number) 
symmetrically  disposed  both  in  position  and  magnitude,  about 
),  we  may  proceed  more  simply,  as  follows :  (Fig.  372). 

From  symmetry  n' 
must  occur  in  the  mid- 
dle of  the  load-line,  of 
which  we  need  lay  off 
only  the  upper  half. 
Take  a  convenient  pole 
0i,  in  the  horizontal 
through  n',  and  draw  a  half  force  diagram  and  a  corres- 
ponding half  equilibrium  polygon  (both  dotted).  The  up- 
per segment  be  of  the  latter  must  be  horizontal  and  being 
prolonged,  cuts  the  prolongation  of  the  first  segment  in  a 
point  d,  which  determines  the  vertical  CD  containing  the 
centre  of  gravity  of  the  loads  occurring  over  the  half -span 
on  the  left.  (See  §  336).  In  the  required  equilibrium  poly- 
gon the  segment  containing  the  point  p  must  be  horizon- 
tal, and  its  intersection  with  the  first  segment  must  lie  in 
CD.  Hence  determine  this  intersection,  C,  by  drawing  the 
vertical  CD  and  a  horizontal  through  p  ;  then  join  CA, 
which  is  the  first  segment  of  the  required  equil.  polygon. 
A  parallel  to  CA  through  1  is  the  first  ray  of  the  corres- 
ponding force  diagram,  and  determines  the  pole  0  on  the 
horizontal  through  n'.  Completing  the  force  diagram  for 


420 


MECHANICS   OF   ENGINEERING. 


this  pole  (half  of  it  only  here),  the  required  equil.  poly- 
gon is  easily  finished  afterwards. 

343.  To  Find  a  System  of  Loads  Under  Which  a  Given  Equi- 
librium Polygon  Would  be  in  Equilibrium.— Fig.  373.     Let  AB 
be  the  given  equilibrium  polygon.     Through  any  point  0 
,      ,  ,      as  a  pole  draw  a  parallel  to  each 

segment  of  the  equilibrium  polygon. 
Any  vertical,  as  V,  cutting  these 
lines  will  have,  intercepted  upon  it, 
a  load-line  1,  2,  3,  whose  parts  1 . .  2, 
2  .  .  3,  etc.,  are  proportional  to  the 
successive  loads  which,  placed  on 
the  corresponding  joints  of  the  equilibrium  polygon  would 
be  supported  by  it  in  equilibrium  (unstable). 

One  load  may  be  assumed  and  the  others  constructed. 
A  hanging,  as  well  as  a  standing,  equilibrium  polygon 
may  be  dealt  with  in  like  manner,  but  will  be  in  stable  equi- 
librium.    The  problem  in  §  44  may  be  solved  in  this  way. 


ARCHES  OF  MASONRY. 


421 


CHAPTER  X. 


RIGHT  ARCHES  OF  MASONRY. 


344. — In  an  ordinary  "right"  stone-arch  (i.e.,  one  in 
which  the  faces  are  "I  to  the  axis  of  the  cylindrical  soffit, 
or  under  surface),  the  successive  blocks  forming  the  arch- 
ring  are  called  voussoirs,  the  joints  between  them  being 
planes  which,  prolonged,  meet  generally  in  one  or  more 
horizontal  lines;  e.g.,  those  of  a  three -centred  arch  in  three 
ii  horizontal  lines ;  those  of  a  circular  arch  in  one,  the  axis 
of  the  cylinder,  etc.  Elliptic  arches  are  sometimes  used.  The 
inner  concave  surface  is  called  the  soffit,  to  which  the  radiat- 
ing joints  between  the  voussoirs  are  made  perpendicular. 
The  curved  line  in  which  the  soffit  is  intersected  by  a  plane 


"1  to  the  axis  of  the  arch  is  the  Intrados.  The  curve  in  the 
same  plane  as  the  intrados,  and  bounding  the  outer  ex- 
tremities of  the  joints  between  the  voussoirs,  is  called  the 
Extrados. 

Fig.  374  gives  other  terms  in  use  in  connection  with  a 


422 


MECHANICS    OF    ENGINEERING. 


stone  arch,  and  explains  those  already  given. 
"  springing-line." 


AS  is  the 


345.  Mortar  and  Friction. — As  common  mortar  hardens 
very  slowly,  no  reliance  should  be  placed  on  its  tenacity 
as  an  element  of  stability  in  arches  of  any  considerable 
size ;  though  hydraulic  mortar  and  thin  joints  of  ordinary 
mortar  can  sometimes  be  depended  on.     Friction,  however, 
between   the   surfaces   of  contiguous   voussoirs,  plays  an 
essential  part  in  the  stability  of  an  arch,  and  will  there- 
fore be  considered. 

The  stability  of  voussoir-arches  must  .*.  be  made  to 
depend  on  the  resistance  of  the  voussoirs  to  compresssion 
and  to  sliding  upon  each  other ;  as  also  of  the  blocks 
composing  the  piers,  the  foundations  of  the  latter  being 
firm. 

346.  Point  of  Application  of  the  Resultant  Pressure  between 
two   consecutive  voussoirs ;   (or  pier  blocks).     Applying 
Navier's  principle  (as  in  flexure  of  beams)  that  the  press- 
ure per  unit  area  on  a  joint  varies  uniformly  from  the 
extremity  under  greatest  compression  to  the  point  of  least 
compression  (or  of  no  compression) ;   and   remembering 
that  negative  pressures  (i.e.,  tension)  can  not  exist,  as  they 
might  in  a  curved  beam,  we  may  represent  the  pressure 
per  unit  area  at  successive  points  of  a  joint  (from  the  intra- 
dos  toward  the  extrados,  or  vice  versa)  by  the  ordinates  of 
a  straight  line,  forming  the  surface  of  a  trapezoid  or  tri- 
angle, in  which  figure  the  foot  of  the  ordinate  of  the  cen- 
tre of  gravity  is  the  point  of  application  of  the  resultant 
pressure.     Thus,  where  the  least  compression  is  supposed 


FIG.  376. 


MASONRY   ARCHES.  423 

to  occur  at  the  intrados  A,  Fig.  375,  the  pressures  vary  as 
the  ordinates  of  a  trapezoid,  increasing  to  a  maximum  value 
at  B,  in  the  extrados.  In  Fig.  376,  where  the  pressure  is  zero 
at  B,  and  varies  as  the  ordinates  of  a  triangle,  the  result- 
ant pressure  acts  through  a  point  one-third  the  joint- 
length  from  A.  Similarly  in  Fig.  377,  it  acts  one-third 
the  joint-length  from  B.  Hence,  when  the  pressure  is  not 
zero  at  either  edge  the  resultant  pressure  acts  within  the 
middle  third  of  the  joint.  Whereas,  if  the  resultant  press- 
ure falls  without  the  middle  third,  it  shows  that  a  portion 
Am  of  the  joint,  see  Fig.  378,  receives  no  pressure,  i.e.,  the 
joint  tends  to  open  along  Am. 

Therefore  that  no  joint  tend  to  open,  the  resultant  press- 
ure must  fall  within  the  middle  third. 

It  must  be  understood  that  the  joint  surfaces  here  dealt 
with  are  rectangles,  seen  edgewise  in  the  figures. 

347.  Friction. — By   experiment   it   has   been   found   the 
angle  of  friction  (see  §  156)  for  two  contiguous  voussoirs 
of  stone  or  brick  is  about  30° ;   i.e.,  the  coefficient  of  fric- 
tion is  /  =  tan.  30°.     Hence  if  the  direction  of  the  press- 
ure exerted  upon  a  voussoir  by  its  neighbor  makes  an 
angle  a  less  than  30°  with  the  normal  to  the  joint  surface, 
there  is  no  danger  of  rupture  of  the  arch  by  the  sliding 
of  one  on  the  other.     (See  Fig.  379). 

348.  Resistance  to  Crushing. — When  the  resultant  pressure 
falls  at  its  extreme  allowable  limit,  viz. :   the  edge  of  the 
middle  third,  the  pressure  per 

unit  of  area  at  nt  Fig.  380,  is 

double  the  mean  pressure  per 

unit  of   area.     Hence,  in  de-  i  * 

signing  an  arch  of  masonry,     N.      / 

we  must  be  assured  that  at        7lf  379 

every  joint  (taking   10   as   a 

factor  of  safety) 

|  Double  the  mean  press-  )  b    legg  than  y 

\  ure  per  unit  of  area         I 


424  MECHANICS   OF   ENGINEERING. 

C  being  the  ultimate  resistance  to  crushing,  of  the  material 
employed  (§  201)  (Modulus  of  Crushing). 

Since  a  lamina  one  foot  thick  will  always  be  considered 
in  what  follows,  careful  attention  must  be  paid  to  the  units 
employed  in  applying  the  above  tests. 

EXAMPLE. — If  a  joint  is  3  ft.  by  1  foot,  and  the  resultant 
pressure  is  22.5  tons  the  mean  pressure  per  sq.  foot  is 

p=22.5-^3=7.5  tons  per  sq.  foot 

.*.  its  double=15  tons  per  sq.  foot=208.3  Ibs.  sq.  inch, 
which  is  much  less  than  J/10  of  C  for  most  building  stones  ; 
see  §  203,  and  below. 

At  joints  where  the  resultant  pressure  falls  at  the  middle, 
the  max.  pressure  per  square  inch  would  be  equal  to  the 
mean  pressure  per  square  inch  ;  but  for  safety  it  is  best  to 
assume  that,  at  times,  (from  moving  loads,  or  vibrations) 
it  may  move  to  the  edge  of  the  middle  third,  causing  the 
max.  pressure  to  be  double  the  mean  (per  square  inch). 

Gen*  Gillmore's  experiments  in  1876  gave  the  following 
results,  among  many  others  : 

NAME  OF  BUILDING  STONE.  C  IN  LBS.  PER  SQ.  INCH. 

Berea  sand-stone,  2-inch  cube,       -        •-  -        -        8955 

4    "        "     -  11720 

Limestone,  Sebastopol,  2-inch  cube  (clialti),  -        -          1075 

Limestone  from  Caen,  France,      -         -  -       .-              3650 

Limestone  from  Kingston,  K  Y.,    -        -  -        -        13900 

Marble,  Vermont,  2-inch  cube,  -       8000  to  13000 

Granite,  New  Hampshire,  2-inch  cube,  15700  to  24000 

349.  The  Three  Conditions  of  Safe  Equilibrium  for  an  arch  of 
uncemented  voussoirs. 

Recapitulating  the  results  of  the  foregoing  paragraphs, 
we  may  state,  as  follows,  the  three  conditions  which  must 
be  satisfied  at  every  joint  of  arch-ring  and  pier,  for  each 
of  any  possible  combination  of  loads  upon  the  structure  : 

(1).  The  resultant  pressure  must  pass  within  the  middle- 
third. 

(2).  The  resultant  pressure  must  not  make  an  angle  > 
30°  with  the  normal  to  the  joint. 

(3).  The  m^an  pressure  per  unit  of  area  on  the  surface 


AKCH    OF    MASONRY. 


425 


of  the  joint  must  not  exceed  l/w  of  the  Modulus  of  crush- 
ing  of  the  material. 

350.  The  True  Linear-Arch,  or  Special  Equilibrium  Polygon; 
and  the  resultant  pressure  at  any  joint.  Let  the  weight 
of  each  voussoir  and  its  load  be  represented  by  a  vertical 
force  passing  through  the  centre  of  gravity  of  the  two,  as 
in  Fig.  381.  Taking  any 
two  points  A  and  B,  A 
being  in  the  first  joint  and 
B  in  the  last ;  also  a  third 
point,  pt  in  the  crown 
joint  (supposing  such  to 
be  there,  although  gener- 
ally a  key-stone  occupies 
the  crown),  through  these 
three  points  can  be  drawn  [§  341]  an  equilibrium  polygon 
for  the  loads  given  ;  suppose  this  equil.  polygon  nowhere 
passes  outside  of  the  arch-ring  (the  arch-ring  is  the  por- 
tion between  the  intrados,  mn,  and  the  (dotted)  extrados 
m'n')  intersecting  the  joints  at  b,  c,  etc.  Evidently  if  such 
be  the  case,  and  small  metal  rods  (not  round)  were  insert- 
ed at  A,  b,  c,  etc.,  so  as  to  separate  the  arch -stones  slight- 
ly, the  arch  would  stand,  though  in  unstable  equilibrium, 
the  piers  being  firm  ;  and  by  a  different  choice  of  A,  p,  and 
B,  it  might  be  possible  to  draw  other  equilibrium  poly- 
gons with  segments  cutting  the  joints  within  the  arch- 
ring,  and  if  the  metal  rods  were  shifted  to  these  new  inter- 
sections the  arch  would  again  stand  (in  unstable  equilib- 
'rium). 

In  other  words,  if  an  arch  stands^  it  may  be  possible  to 
draw  a  great  number  of  linear  arches  within  the  limits  of 
the  arch-ring,  since  three  points  determine  an  equilibrium 
polygon  (or  linear  arch)  for  given  loads.  The  question 
arises  then  :  which  linear  arch  is  the  locus  of  the  actual  re- 
sultant pressures  at  the  successive  joints  ? 

[Considering  the  arch-ring  as  an  elastic  curved  beam 
inserted  in  firm  piers  (i.e.,  the  blocks  at  the  springing-line 


426  MECHANICS    OF    ENGINEERING. 

are  incapable  of  turning)  and  having  secured  a  close  fit  at 
all  joints  before  the  centering  is  lowered,  the  most  satisfac- 
tory answer  to  this  question  is  given  in  Prof.  Greene's 
"  Arches,"  p.  131 ;  viz.,  to  consider  the  arch-ring  as  an 
arch  rib  of  fixed  ends  and  no  hinges ;  see  §  380  of  next 
chapter;  but  the  lengthy  computations  there  employed 
(and  the  method  demands  a  simple  algebraic  curve  for  the 
arch)  may  be  most  advantageously  replaced  by  Prof. 
Eddy's  graphic  method  ("  New  Constructions  in  Graphical 
Statics,"  published  in  Van  Nostrand's  Magazine  for  1877), 
which  applies  to  arch  curves  of  any  form. 

This  method  will  be  given  in  a  subsequent  chapter,  on 
Arch  Bibs,  or  Curved  Beams  ;  but  for  arches  of  masonry  a 
much  simpler  procedure  is  sufficiently  exact  for  practical 
purposes  and  will  now  be  presented]. 

I y         If  two  elastic  blocks 

I         /  \    of  an  arch-ring  touch  at 

~\/^  one  edge>  Fig-  382> tneir 
Jm         adjacent  sides  making  a 
^VSSVN/  small  angle   with  each 

FIG-  382.  FIG.  383.          other,  and  are  then  grad- 

ually pressed  more  and  more  forcibly  together  at  the  edge 
m,  as  the  arch-ring  settles,  the  centering  being  gradually 
lowered,  the  surface  of  contact  becomes  larger  and  larger, 
from  the  compression  which  ensues  (see  Fig.  383);  while 
the  resultant  pressure  between  the  blocks,  first  applied  at 
the  extreme  edge  m,  has  now  probably  advanced  nearer  the 
middle  of  the  joint  in  the  mutual  adjustment  of  the  arch- 
stones.  With  this  in  view  we  may  reasonably  deduce  the 
following  theory  of  the  location  of  the  true  linear  arch 
(sometimes  called  the  "  line  of  pressures  "  and  "  curve  of 
pressure")  in  an  arch  under  given  loading  and  with^rw 
piers.  (Whether  the  piers  are  really  unyielding,  under  the 
oblique  thrusts  at  the  springing-line,  is  a  matter  for  sub- 
sequent investigation. 

351.  Location  of  the  True  Linear  Arch. — Granted  that  the- 
voussoirs  have  been  closely  fitted  to  each  other  over  the 


ARCH   OF   MASONRY.  427 

centering  (sheets  of  lead  are  sometimes  used  in  the  joints 
to  make  a  better  distribution  of  pressure);  and  that  the 
piers  are  firm  ;  and  that  the  arch  can  stand  at  all  without 
the  centering  ;  then  we  assume  that  in  the  mutual  accom- 
modation between  the  voussoirs,  as  the  centering  is  low- 
ered, the  resultant  of  the  pressures  distributed  over  any 
joint,  if  at  first  near  the  extreme  edge  of  the  joint,  advances 
nearer  to  the  middle  as  the  arch  settles  to  its  final  posi- 
tion of  equilibrium  under  its  load ;  and  hence  the  follow- 
ing 

352.  Practical  Conclusions. 

I.  If  for  a  given  arch  and  loading,  with  firm   piers,  an 
equilibrium  polygon  can  be  drawn  (by  proper  selection  of 
the  points  A,  p,  and  B,  Fig.  381)  entirely  within  the  mid- 
dle third  of  the  arch  ring,  not  only  will  the  arch  stand,  but 
the  resultant  pressure  at  every  joint  will  be  within  the 
middle  third  (Condition  1,  §  349) ;  and  among  all  possible 
equilibrium  polygons  which  can  be  drawn  within  the  mid- 
dle third,  that  is  the  "  true  "  one  which  most  nearly  coin- 
cides with  the  middle  line  of  the  arch-ring. 

II.  If  (with  firm  piers,  as  before)  no  equilibrium  poly- 
gon can  be  drawn  within  the  middle  third,  and  only  one 
within  the  arch-ring  at  all,  the  arch  may  stand,  but  chip- 
ping and  spawling  are  likely  to  occur  at  the  edges  of  the 
joints.     The  design  should  .-.  be  altered. 

III.  If  no  equilibrium  polygon  can  be  drawn  within 
the  arch-ring,  the  design  of  either  the  arch  or  the  loading 
must   be    changed  ;  since,  although  the  arch  may  stand, 
from  the  resistance  of  the  spandrel  walls,  such  a  stability 
must  be  looked  upon  as  precarious  and  not  countenanced 
in  any   large  important  structure.     (Very  frequently,  in 
small  arches  of  brick  and  stone,  as  they  occur  in  buildings, 
the  cement  is  so  tenacious  that  the  whole  structure  is  vir- 
tually a  single  continuous  mass). 

When  the  "  true  "  linear  arch  has  once  been  determined,, 
the  amount  of  the  resultant  pressure  on  any  joint  is  given, 
by  the  length  of  the  proper  ray  in  the  force  diagram. 


428 


MECHANICS   OF   ENGINEERING. 


ARRANGEMENT  OF  DATA  FOR  GRAPHIC 
TREATMENT. 

353.  Character  of  Load. — In  most  large  stone  arch  bridges 
the  load  (permanent  load)  does  not  consist  exclusively  of 
masonry  up  to  the  road-way  but  partially  of  earth  filling 
above  the  masonry,  except  at  the  faces  of  the  arch  where 
the  spandrel  walls  serve  as  retaining  walls  to  hold  the 
earth.  (Fig.  384).  If  the  intrados  is  a  half  circle  or  half- 


FIG.  384. 


ellipse,  a  compactly -built  masonry  backing  is  carried  up 
beyond  the  springing-line  to  AB  about  60°  to  45°  from  the 
crown,  Fig.  385 ;  so  that  the  portion  of  arch  ring  below 
AB  may  be  considered  as  part  of  the  abutment,  and  thus 
AB  is  the  virtual  springing-line,  for  graphic  treatment. 

Sometimes,  to  save  filling,  small  arches  are  built  over 
the  haunches  of  the  main  arch,  with  earth  placed  over 
them,  as  shown  in  Fig.  386.  In  any  of  the  preceding  cases 


it  is  customary  to  consider  that,  on  account  of  the  bond- 
ing of  the  stones  in  the  arch  shell,  the  loading  at  a  given 
distance  from  the  crown  is  uniformly  distributed  over  the 
width  of  the  roadway. 


ARCHES   OF   MASONRY.  429 

354.  Reduced  Load-Contour.— In  the  graphical  discussion 
of  a  proposed  arch  we  consider  a  lamina  one  foot  thick, 
this  lam  iiia  being  vertical  and  ~]  to  the  axis  of  the  arch  ; 
i.e.,  the  lamina  is  ||  to  the  spandrel  walls.     For  graphical 
treatment,  equal  areas  of  the  elevation  (see  Fig.  387)  of 
this  lamina  must  represent  equal   weights.     Taking  the 
material  of  the  arch-ring  as  a  standard,  we  must  find  for 
each  point  p  of  the  extrados  an  imaginary  height  z  of  the 
arch-ring  material,  which  would  give  the  same  pressure 
(per  running  horizontal  foot)  at  that  point  as  that  due  to 
the  actual  load  above  that  point.     A  number  of  such  or- 
dinates,  each  measured  vertically  upward  from   the  extra- 
dos  determine   points  in  the  "Reduced   Load-Contour,"  i.e., 
the  imaginary  line,  AM,  the  area  between  which   and  the 
extrados  of  the  arch -ring  represents  a  homogeneous  load 
of  the  same  density  as  the  arch-ring,  and  equivalent  to  the 
actual  load  (above  extrados),  vertical  by  vertical. 

355.  Example  of  Reduced  Load-Contour.— Fig.    388.     Given 
an   arch-ring  of  granite  (heaviness  =  170  Ibs.  per  cubic 
foot)  with  a  dead  load  of  rubble  (heav.  =  140)  and  earth 
(heav.  =  100),  distributed  as  in  figure.     At  the  point  p,  of 
the  extrados,  the  depth  5  feet  of  rubble  is  equivalent  to  a 
depth  of  [|*  x5]=4.1  ft.  of  granite,  while  the  6  feet  of  earth 
is  equivalent  to  [j^x6]=3.5   feet  of  granite.     Hence  the 
Reduced  Load-Contour  has  an  ordinate,  above  p,  of   7.6  feet. 
That  is,  for  each  of  several  points  of  the  arch -ring  extrados 
reduce  the  rubble  ordinate  in  the  ratio  of  170  :  140,  and 
the  earth  ordinate  in  the  ratio  170  :  100  and  add  the  re- 
sults, setting  off  the  sum  vertically  from  the  points  in  the 
extrados*.     In  this  way  Fig.  389  is  obtained  and  the  area 

*This  Is  most  conveniently  done  by  graphics,  thus  :  On  a  right-line  set  off  17  equal 
parts  (of  any  convenient  magnitude.)  Call  this  distance  OA.  Through  0  draw  another 
right  line  at  any  convenient  angle  (30°  to  60°)  with  OA,  and  on  it  from  O 

set  off  OB  equal  to  14  (for  the  rubble  ;  or  10  for  the  earth)  of  the  same  equal 
parts.  Join  AB.  From  O  toward  A  set  off*  all  the  rubble  ordinates  to  be  reduced, 
(each  being  set  off  from  0)  and  through  the  other  extremity  of  each  draw  a  line  par- 
allel to  AB.  The  reduced  ordinates  will  be  the  respective  lengths,  from  0,  along  OB, 
to  the  intersections  of  these  parallels  with  OB. 

*  With  the  dividers. 


430 


MECHANICS   OF   ENGINEERING. 


there  given  is  to  be  treated  as  representing  homogeneous 
granite  one  foot  thick.  This,  of  course,  now  includes  the 
arch-ring  also.  AB  is  the  "  reduced  load-contour." 

356.  Live  Loads. — In  discussing  a  railroad  arch  bridge 
the  "  live  load  "  (a  train  of  locomotives,  e.g.,  to  take  an  ex- 
treme case)  can  not  be  disregarded,  and  for  each  of  its  po- 
sitions we  have  a  separate  Reduced  Load-Contour. 

EXAMPLE. — Suppose  the  arch  of  Fig.  388  to  be  12  feet 
wide  (not  including  spandrel  walls)  and  that  a  train  of  lo- 
comotives weighing  3,000  Ibs.  per  running  foot  of  the  track 
covers  one  half  of  the  span.  Uniformly  *  distributed  later- 
ally over  the  width,  12  ft.,  this  rate  of  loading  is  equiva- 
lent to  a  masonry  load  of  one  foot  high  and  a  heaviness  of 
250  Ibs.  per  cubic  ft.,  i.e.,  is  equivalent  to  a  height  of  1.4 
ft.  of  granite  masonry  [since  ^  x  1.0— 1.4]  over  the  half 
span  considered.  Hence  from  Fig.  390  we  obtain  Fig.  391 
in  an  obvious  manner.  Fig.  391  is  now  ready  for  graphic 
treatment. 


FIG.  391. 

357.  Piers  and  Abutments. — In  a  series  of  equal  arches 
the  pier  between  two  consecutive  arches  bears  simply  the 
weight  of  the  two  adjacent  semi-arches,  plus  the  load  im- 

*  If  the  earth-filling  is  sLallcw,  the  Ir.minae  directly  under  the  track  prob- 
ably receive  a  greater  pressure  than  the  others. 


ARCHES   OF  MASONRY. 


431 


mediately  above  the  pier,  and  .•.  does  not  need  to  be  as 
large  as  the  abutment  of  the  first  and  last  arches,  since 
these  latter  must  be  prepared  to  resist  the  oblique  thrusts 
of  their  arches  without  help  from  the  thrust  of  another  on 
the  other  side. 

'  In  a  very  long  series  of  arches  it  is  sometimes  customary 
to  make  a  few  of  the  intermediate  piers  large  enough  to 
act  as  abutments.  These  are  called  "  abutment  piers,"  and 
in  case  one  arch  should  fall,  no  others  would  be  lost  except 
those  occurring  between  the  same  two  abutment  piers  as 
the  first.  See  Fig.  392.  A  is  an  abutment-pier. 


nn-nnn 


GRAPHICAL,  TREATMENT  OF  ARCH. 

358. — Having  found  the  "  reduced  load-contour,"  as  in 
preceding  paragraphs,  for  a  given  arch  and  load,  we  are 
ready  to  proceed  with  the  graphic  treatment,  i.e.,  the  first 
given,  or  assumed,  form  and  thickness  of  arch-ring  is  to  be 
investigated  with  regard  to  stability.  It  may  be  necessary 
to  treat,  separately,  a  lamina  under  the  spandrel  wall,  and 
one  under  the  interior  loading.  The  constructions  are 
equally  well  adapted  to  arches  of  all  shapes,  to  Gothic  as 
well  as  circular  and  elliptical. 

359.— Case  I  Symmetrical  Arch  and  Symmetrical  Loading. — 
(The  "  steady  "  (permanent)  or  "  dead  "  load  on  an  arch  is 
usually  symmetrical).  Fig.  393.  From  symmetry  we  need 


432  MECHANICS   OF   ENGINEERING. 

deal  with  only  one  half  (say  the  left)  of  the  arch  and  load. 
Divide  this  semi-arch  and  load  into  six  or  ten  divisions 
by  vertical  lines ;  these  divisions  are  considered  as  trape- 
zoids  and  should  have  the  same  horizontal  width  =  b  (a 
convenient  whole  number  of  feet)  except  the  last  one,  LKN, 
next  the  abutment,  and  this  is  a  pentagon  of  a  different 
width  ft,,  (the  remnant  of  the  horizontal  distance  LC).  The 
weight  of  masonry  in  each  division  is  equal  to  (the  area 
of  division)  x  (unity  thickness  of  lamina)  x  (weight  of  a  cu- 
bic unit  of  arch-ring).  For  example  for  a  division  having 
an  area  of  20  sq.  feet,  and  composed  of  masonry  weighing 
160  Ibs.  per  cubic  foot,  we  have  20x1x160—3,200  Ibs., 
applied  through  the  centre  of  gravity  of  the  division. 
The  area  of  a  trapezoid,  Fig.  394,  is  ^&(Ai+^2)>  and  its  cen- 
tre of  gravity  may  be  found,  Fig.  395,  by  the  construction 
of  Prob.  6,  in  §  26 ;  or  by  §  27«.  The  weight  of  the  pen- 
tagon LN,  Fig.  393,  and  its  line  of  application  (through 
centre  of  gravity)  may  be  found  by  combining  results  for 
the  two  trapezoids  into  which  it  is  divided  by  a  vertical 
through  K.  See  §  21. 

Since  the  weights  of  the  respective  trapezoids  (except, 
ing  LN)  are  proportional  to  their  middle  vertical  in- 
tercepts [such  as  ^(^1+^2)  Fig.  394]  these  intercepts  (trans- 
ferred with  the  dividers)  may  be  used  directly  to  form  the 
load-line,  Fig.  396,  or  proportional  parts  of  them  if  more 
convenient.  The  force  scale,  which  this  implies,  is  easily 
computed,  and  a  proper  length  calculated  to  represent  the 
weight  of  the  odd  division  LN ;  i.e.,  1  ...  2  on  the  load- 
line. 

Now  consider  A,  the  middle  point  of  the  abutment  joint, 
Fig.  396,  as  the  starting  point  of  an  equilibrium  polygon 
(or  abutment  of  a  linear  arch)  for  a  given  loading,  and  re- 
quire that  this  equilibrium  polygon  shall  pass  through  p, 
the  middle  of  the  crown  joint,  and  through  the  middle  of 
the  abutment  joint  on  the  right  (not  shown  in  figure). 

Proceed  as  in  §  342,  thus  determining  the  polygon  Ap 
for  the  half-arch.  Draw  joints  in  the  arch-ring  through 
those  points  where  the  extrados  is  intersected  by  the  ver- 


AKCIIES   OF   MASONBY. 


433 


FIG.  390.  FIG.  397. 

tical  separating  the  divisions  (not  the  gravity  verticals). 
The  points  in  which  these  joints  are  cut  by  the  segments 
of  the  equilibrium  polygon,  Fig.  397,  are  (very  nearly,  if 
the  joint  is  not  more  than  60°  from  p,  the  crown)  the  points 
of  application  in  these  joints,  respectively,  of  the  resultant 
pressures  on  them,  (if  this  is  the  "  true  linear  arch  "  for 
this  arch  and  load)  while  the  amount  and  direction  of  each 
such  pressure  is  given  by  the  proper  ray  in  the  force -dia- 
gram. 

If  at  any  joint  so  drawn  the  linear  arch  (or  equilibrium 
polygon)  passes  outside  the  middle  third  of  the  arch-ring, 
the  point  A,  or  p,  (or  both)  should  be  judiciously  moved 
(within  the  middle  third)  to  find  if  possible  a  linear  arch 
which  keeps  within  limits  at  all  joints.  If  this  is  found 
impossible,  the  thickness  of  the  arch -ring  may  be  increased 
at  the  abutment  (giving  a  smaller  increase  toward  the 
crown)  and  the  desired  result  obtained  ;  or  a  change  in  the 
distribution  or  amount  of  the  loading,  if  allowable,  may 
gain  this  object.  If  but  one  linear  arch  can  be  drawn 
within  the  middle  third,  it  may  be  considered  the  "  true  " 
one ;  if  several,  the  one  most  nearly  co-inciding  with  the 
middles  of  the  joints  (see  §§  351,and  352)  is  so  considered. 

360.— Case  II.  Unsymmetrical  Loading  on  a  Symmetrical  Arch; 
(e.g.,  arch  with  live  load  covering  one  half -span  as  in  Figs. 
390  and  391).  Here  we  must  evidently  use  a  full  force 
diagram,  and  the  full  elevation  of  the  arch -ring  and  load. 


434 


MECHANICS  OF    ENGINEERING. 


See  Fig.  398.      Select  three  points  A,  p,  and  B,  as  follows, 
to  determine  a  trial  equilibrium  polygon : 

Select  A  at  the  loiccr  limit  of  the  middle  third  of  the 


abutment-joint  at  the  end  of  the  span  which  is  the  more 
heavily -loaded  ;  in  the  other  abutment-joint  take  B  at  the 
upper  limit  of  the  middle  third ;  and  take  p  in  the  middle 
of  the  crown-joint.  Then  by  §  341  draw  an  equilibrium 
polygon  (i.e.,  a  linear  arch)  through  these  three  points  for 
the  given  set  of  loads,  and  if  it  does  not  remain  within  the 
middle  third,  try  other  positions  for  A,  p,  and  J5,  within 
the  middle  third.  As  to  the  "  true  linear  arch  "  alterations 
of  the  design,  etc.,  the  same  remarks  apply  as  already 
given  in  Case  I.  Very  frequently  it  is  not  necessary  to 
draw  more  than  one  linear  arch,  for  a  given  loading,  for 
even  if  one  could  be  drawn  nearer  the  middle  of  the  arch- 
ring  than  the  first,  that  fact  is  most  always  apparent  on 
mere  inspection,  and  the  one  already  drawn  (if  within 
middle  third)  will  furnish  values  sufficiently  accurate  for 
the  pressures  on  the  respective  joints,  and  their  direction 
angles. 

360a. — The  design  for  the  arch-ring  and  loading  is  not 
to  be  considered  satisfactory  until  it  is  ascertained  that  for 
the  dead  load  and  any  possible  combination  of  live-load 
'in  addition)  the  pressure  at  any  joint  is 


ARCHES   OF   MASONRY.  435 

(1.)    Within  the  middle  third  of  that  joint ; 

(2.)  At  an  angle  of  <  30°  with  the  normal  to  joint- 
surface. 

(3.)  Of  a  mean  pressure  per  square  inch  not  >  than  l/x 
of  the  ultimate  crushing  resistance.  (See  §  348.) 

§361.  Abutments. — The  abutment  should  be  compactly 
and  solidly  built,  and  is  then  treated  as  a  single  rigid  mass. 
The  pressure  of  the  lowest  voussoir  upon  it  (considering 
a  lamina  one  foot  thick)  is  given  by  the  proper  ray  of  the 
force  diagram  (0  ..  l,e.  g.,  in  Fig.  396)  in  amount  and  direc- 
tion. The  stability  of  the  abutment  will  depend  on  the 
amount  and  direction  of  the  resultant  obtained  by  com- 
bining that  pressure  Pa  with  the  weight  G  of  the  abutment 
and  its  load,  see  Fig.  399.  Assume  a  probable  width  RS 
for  the  abutment  and  compute  the  weight  G 
of  the  corresponding  abutment  OBRS  and 
MNBO,  and  find  the  centre  of  gravity  of  the 
whole  mass  C.  Apply  G  in  the  vertical 
through  (7,  and  combine  it  with  Pa  at  their  in- 
tersection D.  The  resultant  P  should  not  cut 
the  base  It  Sin  a  point  beyond  the  middle  third 
(or,  if  this  rule  gives  too  massive  a  pier,  take 
such  a  width  that  the  pressure  per  square 
inch  at  S  shall  not  exceed  a  safe  value  as 
FIG.  399.  computed  from  §  362.)  After  one  or  two 
trials  a  satisfactory  width  can  be  obtained. 
We  should  also  be  assured  that  the  angle  PDG  is  less 
than  30°.  The  horizontal  joints  above  ES  should  also  be 
tested  as  if  each  were,  in  turn,  the  lowest  base,  and  if 
necessary  may  be  inclined  (like  mn)  to  prevent  slipping. 
On  no  joint  should  the  maximum  pressure  per  square  inch 
be  >  than  l/m  the  crushing  strength  of  the  cement.  Abut- 
ments of  firm  natural  rock  are  of  course  to  be  preferred 
where  they  can  be  had.  If  water  penetrates  under  an 
abutment  its  buoyant  effort  lessens  the  weight  of  the  lat- 
ter to  a  considerable  extent. 


436 


MECHANICS   OF   ENGINEEEING. 


362.  Maximum  Pressure  Per  Unit  of  Area  When  the  Resultant 
Pressure  Falls  at  Any  Given  Distance  from  the  Middle ;  according 
to  Navier's  theory  of  the  distribution  of  the  pressure  ;  see 
§  346.  Case  I.  Let  the  resultant  pressure  P,  Fig.  400,  (a), 


(a) 


FIG.  401. 


fall  within  the  middle  third,  a  distance  =  nd  (  <  %  d) 
from  the  middle  of  joint  (d  —  depth  of  joint.)  Then  we 
have  the  following  relations  : 

p  (the  mean  press,  per.  sq.  in.),/?,,,  (max.  press,  persq.  in.), 
and  pa  (least  press,  per  sq.  in.)  are  proportional  to  the  lines 
h  (mid.  width),  a  (max.  base),  and  c  (min.  base)  respectively, 
of  a  trapezoid,  Fig.  400,  (b),  through  whose  centre  of  gravity 
P  acts.  But  (§  26) 


6  a+c 
•'•  Pm=P  (6»+l).     Hence  the  following  table  : 


=  *6  d 


press.  pm=   2 


V.d 


7m  d 


then  the  max. 

times  the  mean  pressure. 


Case  II.  Let  P  fall  outside  the  mid.  third,  a  distance = 
nd  (>  l/6  d)  from  the  middle  of  joint.  Here,  since  the 
joint  is  not  considered  capable  of  withstanding  tension, 
we  have  a  triangle,  instead  of  a  trapezoid.  Fig.  401.  First 
compute  the  mean  press,  per  sq.  in. 


or   from   this  table:    (lamina  one 


(l-2n)  18  d  inches 
foot  thick). 


ARCHES    OF   MASONRY. 


437 


For  nd  = 

A<* 

&d 

^d 

ft* 

•&<* 

A^ 

P  = 

i     p 

10  '  d 

1      P 

8   '  d 

1       P 
6   '  d 

1      P 

4  '  d 

1      P 

2   '  d 

infinity. 

(d  in  inches  and  Pin  Ibs. ;  with  arch  lamina  1  ft.  in  thickness.) 

Then  the  maximum  pressure  (at  A,  Fig.  401)  pm,  =  %p, 
becomes  known,  in  Ibs.  per  sq.  in. 

362a.  Arch-ring  under  Non-vertical  Forces. — An  example  of 
this  occurs  when  a  vertical  arch-ring  is  to  support  the  pressure 
of  a  liquid  on  its  extrados.  Since  water-pressures  are  always 
at  right  angles  to  the  surface  pressed  on,  these  pressures  on  the 
extradosal  surface  of  the  arch- ring  form  a  system  of  non-paral- 
lel forces  which  are  normal  to  the  curve  of  the  extrados  at 
their  respective  points  of  application  and  lie  in  parallel 
vertical  planes,  parallel  to  the  faces  of  the  lamina.  We  here 
assume  that  the  extradosal  surface  is  a  cylinder  (in  the  most 
general  sense)  whose  rectilinear  elements  are  ~|  to  the  faces  of 
the  lamina.  If,  then,  we  divide  the  length  of  the  extrados, 
from  crown  to  each  abutment,  into  from  six  to  ten  parts,  the 
respective  pressures  on  the  corresponding  surfaces  are  obtained 
by  multiplying  the  area  of  each  by  the  depth  of  its  centre  of 
gravity  from  the  upper  free  surface  of  the  liquid,  and  this 
product  by  the  weight  of  a  unit  of  volume  of  the  liquid  ;  and 
each  such  pressure  may  be  considered  as  acting  through  the 
centre  of  the  area.  Finally,  if  we  find  the  resultant  of  each 
of  these  pressures  and  the  weight  of  the  corresponding  portion 
of  the  arch-ring,  these  resultants  form  a  series  of  non-vertical 
forces  in  a  plane,  for  which  an  equilibrium  polygon  can  then 
be  passed  through  three  assumed  points  by  §  3Y8a,  these  three 
points  being  taken  in  the  crown-joint  and  the  two  abutment- 
joints.  As  to  the  "  true  linear  arch"  see  §  359. 

As  an  extreme  theoretic  limit  it  is  worth  noting  that  if  the 
extrados  and  intrados  of  the  arch-ring  are  concentric  circles;  if 
the  weights  of  the  voussoirs  are  neglected ;  and  if  the  rise  of 
the  arch  is  very  small  compared  with  the  depth  of  the  crown 
below  the  water  surface,  then  the  circularcentre-line  of  the 
arch-ring  is  the  "  true  linear  arch" 


438  MECHANICS   OF   ENGINEERING. 


CHAPTEE  XL 


ARCH-RIBS. 


364.  Definitions  and  Assumptions. — An  arch-rib   (or  elastic- 
arch,  as  distinguished  from  a  block-work  arch)  is  a  rigid 
curved  beam,   either   solid,  or  built  up  of  pieces  like  a 
truss  (and  then  called  a  braced  arch)  the  stresses  in  which, 
under  a  given  loading  and  with  prescribed  mode  of  sup- 
port it  is  here  proposed  to  determine.     The  rib  is  sup- 
posed symmetrical  about  a  vertical  plane  containing  its 
axis  or  middle  line,  and  the  Moment  of  Inertia  of  any  cross 
section  is  understood  to  be  referred  to  a  gravity  axis  of 
the  section,  which  (the  axis)  is  perpendicular  to  the  said 
vertical  plane.     It  is  assumed  that  in  its  strained  condi- 
tion under  a  load,  the  shape  of  the  rib  differs  so  little 
from  its  form  when  unstrained  that  the  change  in  the  ab- 
scissa or  ordinate  of  any  point  in  the  rib  axis  (a  curve) 
may  be  neglected  when  added  (algebraically)  to  the  co- 
ordinate itself ;  also  that  the  dimensions  of  a  cross-section 
are  small  compared  with  the  radius  of  curvature  at  any 
part  of  the  curved  axis,  and  with  the  span. 

365.  Mode  of  Support. — Either  extremity  of  the  rib  may  be 
hinged  to  its  pier  (which  gives  freedom  to  the  end-tangent- 
line  to  turn  in  the  vertical  plane  of  the  rib  when  a  load  is 
applied);  or  may  \yejixed,  i.e.,  so  built-in,  or  bolted  rigid- 
ly to  the  pier,  that  the  end-tangent-line  is  incapable  of 
changing  its  direction  when  a  load  is  applied.     A  hinge 
may  be  inserted  anywhere  along  the  rib,  and  of  course 


ARCH   RIBS. 


439 


destroys  the  rigidity,  or  resistance  to  bending  at  that 
point.  (A.  hinge  having  its  pin  horizontal  "|  to  the  axis  of 
the  rib  is  meant).  Evidently  no  more  than  three  such 
hinges  could  be  introduced  along  an  arch- rib  between  two 
piers  ;  unless  it  is  to  be  a  hanging  structure,  acting  as  a 
suspension-cable. 

366.  Arch  Rib  as  a  Free  Body. — In  considering  the  whole 
rib  free  it  is  convenient,  for  graphical  treatment,  that  no 
section  be  conceived  made  at  its  extremities,  if  fixed  ;  hence 
in  dealing  with  that  mode  of  support  the  end  of  the  rib 
will  be  considered  as  having  a  rigid  prolongation  reach- 
ing to  a  point  vertically  above  or  below  the  pier  junction, 
an  unknown  distance  from  it,  and  there  acted  on  by  a  force 
of  such  unknown  amount  and  direction  as  to  preserve  the 
actual  extremity  of  the  rib  and  its  tangent  line  in  the  same 
position  and  direction  as  they  really  are.  As  an  illustra- 
tion of  this  Fig.  402 
shows  free  an  arch  rib. 
ONB,  with  its  extremi- 
ties 0  and  B  fixed  in  the 
piers,  with  no  hinges,  Q\ 
and  bearing  two 
loads  P.  and  P2.  The 
other  ces  of  the  sys- 
tem holding  it  in  equi- 
librium are  the  horizontal  and  vertical  components,  of  the 
pier  reactions  (Ht  V,  HUJ  and  Vn),  and  in  this  case  of  fixed 
ends  each  of  these  two  reactions  is  a  single  force  not  in- 
tersecting the  end  of  the  rib,  but  cutting  the  vertical 
through  the  end  in  some  point  F  (on  the  left ;  and  in  G  on 
the  right)  at  some  vertical  distance  c,  (or  d\  from  the  end. 
Hence  the  utility  of  these  imaginary  prolongations  OQF, 
and  BRG,  the  pier  being  supposed  removed.  Compare 
Figs.  348  and  350. 

The  imaginary  points,  or  hinges,  F  and  G,  will  be  called 
abutments  being  such  for  the  special  equilibrium  polygon 


FIG.  402. 


440  MECHANICS  OF  ENGINEERING. 

(dotted  line),  while  0  and  B  are  the  real  ends  of  the  curved 
beam,  or  rib. 

In  this  system  of  forces  there  are  five  unknowns,  viz.:  Vt 
VM  B  —  H0,  and  the  distances  c  and  d.  Their  determina- 
tion by  analysis,  even  if  the  rib  is  a  circular  arc,  is  ex- 
tremely intricate  and  tedious  ;  but  by  graphical  statics 
(Prof.  Eddy's  method ;  see  §  350  for  reference),  it  is  com- 
paratively simple  and  direct  and  applies  to  any  shape  of 
rib,  and  is  sufficiently  accurate  for  practical  purposes. 
This  method  consists  of  constructions  leading  to  the  loca- 
tion of  the  "  special  equilibrium  polygon "  and  its  force 
diagram.  In  case  the  rib  is  hinged  to  the  piers,  the  re- 
actions of  the  latter  act  through  these  hinges,  Fig.  403, 
i.e.,  the  abutments  of  the  special 
equilibrium  polygon  coincide  with 
the  ends  of  the  rib  0  and  B,  and  for 
a  given  rib  and  load  the  unknown 
quantities  are  only  three  V,  Vllt  and 
H\  (strictly  there  are  four  ;  but  IX 
=  0  gives  Ha  =  H).  The  solution  FlG.  403. 

by  analytics  is  possible  only  for  ribs  of  simple  algebraic 
curves  and  is  long  and  cumbrous ;  whereas  Prof.  Eddy's 
graphic  method  is  comparatively  brief  and  simple  and  is 
applicable  to  any  shape  of  rib  whatever. 

367.  Utility  of  the  Special  Equilibrium  Polygon  and  its  force 
diagram.  The  use  of  locating  these  will  now  be  illustrated 
[See  §  332].  As  proved  in  §§  332  and  334  the  compres- 
sion in  each  "  rod  "  or  segment  of  the  "  special  equilibrium 
polygon"  is  the  anti-stress  resultant  of  the  cross  sections  in 
the  corresponding  portion  of  the  beam,  rib,  or  other  struc- 
ture, the  value  of  this  compression  (in  Ibs.  or  tons)  being 
measured  by  the  length  of  the  parallel  ray  in  the  force 
diagram.  Suppose  that  in  some  way  (to  be  explained  sub- 
sequently) the  special  equilibrium  polygon  and  its  force 
diagram  have  been  drawn  for  the  arch-rib  in  Fig.  404  hav- 
ing fixed  ends,  0  and  B,  and  no  hinges  ;  required  the  elastic 
stresses  in  any  cross-section  of  the  rib  as  at  ra.  Let  the 


ARCH   KIBS. 


441 


H-" 

FiG.  404. 

scale  of  the  force-diagram  on  the  right  be  200  Ibs.  to  the 
inch,  say,  and  that  of  the  space -diagram  (on  the  left)  30  ft. 
to  the  inch. 

The  cross  section  m  lies  in  a  portion  TK,  of  the  rib,  cor- 
responding to  the  rod  or  segment  be  of  the  equilibrium 
polygon;  hence  its  anti-stress-resultant  is  a  force  R2  acting 
in  the  line  6c,  and  of  an  amount  given  in  the  force-diagram. 
Now  R2  is  the  resultant  of  V,  H,  and  Plf  which  with  the 
elastic  forces  at  m  form  a  system  in  equilibrium,  shown  in 
Fig.  405  ;  the  portion  FO  Tm  being  considered  free.  Hence 


Pis.  405. 


taking  the  tangent  line  and  the  normal  at  m  as  axes  we 
should  have  Z  (tang,  comps.)  =  0  ;  I  (norm,  comps.)  =  0  ; 
and  I  (moms,  about  gravity  axis  of  the  section  at  ra)  =  0, 
and  could  thus  find  the  unknowns  plt  p.2,  and  J",  which  ap- 
pear in  the  expressions  piF  the  thrust,  ^--  the  moment  of 


442  MECHANICS   OF   ENGINEERING. 

the  stress-couple,  and  J"the  shear.  These  elastic  stresses 
are  classified  as  in  §  295,  which  see.  py  and  p2  are  Ibs.  per 
square  inch,  e/is  Ibs.,  e  is  the  distance  from  the  horizontal 
gravity  axis  of  the  section  to  the  outermost  element  of 
area,  (where  the  compression  or  tension  is  p2  Ibs.  per  sq. 
in.,  as  due  to  the  stress-couple  alone)  while  1  is  the  "  mo- 
ment of  inertia  "  of  the  section  about  that  gravity  axis. 
[See  §§  247  and  295  ;  also  §  85].  Graphics,  however,  gives 
us  a  more  direct  method,  as  follows  :  Since  R2,  in  the  line 
be,  is  the  equivalent  of  V,  H,  and  Plt  the  stresses  at  m  will 
be  just  the  same  as  if  R2  acted  directly  upon  a  lateral  pro- 
longation of  the  rib  at  T  (to  intersect  ftcFig.  405)  as  shown 
in  Fig.  406,  this  prolongation  Tb  taking  the  place  of  TOF 
in  Fig.  405.  The  force  diagram  is  also  reproduced  here. 
Let  a  denote  the  length  of  the  "|  from  ra's  gravity  axia 
upon  be,  and  z  the  vertical  intercept  between  m  and  be. 
For  this  imaginary  free  body,  we  have, 


from  -  (tang.  compons.^O,  R,  cos  a= 

and  from  2  (norm.  compons.)=0,JR2  sin  a=J 

while  from  -T(  moms,  about)  )         -,          „          p2I 
the  gravity  axis  of  m)=0,      \  We  have  **  =  ~e  ' 

But  from  the  two  similar  triangles  (shaded  ;  one  of  them 
is  in  force  diagram)  a  :z  ::  H:R,  .'.  R.2a=Hz,  whence  we 
may  rewrite  these  relations  as  follows  (with  a  general  state- 
ment), viz.: 

If  the  Special  Equilibrium  Polygon  and  Its  Force  Diagram  Have 
Been  Drawn  for  a  given  arch-rib,  of  given  mode  of  support, 
and  under  a  given  loading,  then  in  any  cross-section  of  the 
rib,  we  have  (F  =  area  of  section): 

The  projection  of  the  proper 

/i  \     mu    rm,  of  tlie  force 

(1.)    The  Thrust, 


f  tho 
drawn  at  the  given  section. 


ARCH    RIBS.  443 

(2.)     The  Shear,  i.e.,  J,  = 

(upon  which  dependsthe        The  projection  of  the  proper 
ray  (of  the  force  diagram)  up- 


'3.)     The  Moment  of   the 
stress  couple,  i.e.,  £?-  ,  = 


on  the  normal  to  the  rib  curve 


shearing    stress   in   the 

web).     (See  §§  253  and 

'      v  at  the  given  section. 

256). 

The  product  (Hz)  of  the  H 
(or  pole-distance)  of  the  force- 
diagram  by  the  vertical  dis- 
tance of  the  gravity  axis  of  the 
section  from  the  spec,  equilib- 
rium polygon. 

By  the  "  proper  ray  "  is  meant  that  ray  which  is  parallel 
to  the  segment  (of  the  equiL  polygon)  immediately  under 
or  above  which  the  given  section  is  situated.  Thus  in 
Fig.  404,  the  proper  ray  for  any  section  on  TK  is  B2 ;  on 
KB,  RS ;  on  TO,  Bv.  The  projection  of  a  ray  upon  any 
given  tangent  or  normal,  is  easily  found  by  drawing  through 
each  end  of  the  ray  a  line  ~|  to  the  tangent  (or  normal) ; 
the  length  between  these  ~]  's  on  the  tangent  (or  normal)  is 
the  force  required  (by  the  scale  of  the  force  diagram).  We 
may  thus  construct  a  shear  diagram,  and  a  thrust  diagram 
for  a  given  case,  while  the  successive  vertical  intercepts 
between  the  rib  and  special  equilibrium  polygon  form  a 
moment  diagram.  For  example  of  the  z  of  a  point  m  is  ^ 
inch  in  a  space  diagram  drawn  to  a  scale  of  20  feet  to  the 
inch,  while  H  measures  2.1  inches  in  a  force  diagram  con- 
structed on  a  scale  of  ten  tons  to  the  inch,  we  have,  for  the 
moment  of  the  stress-couple  at  ra,  M =Hz=  [2.1  X 10]  tons 
X[^x20]ft.=210ft.  tons. 

368. — It  is  thus  seen  how  a  location  of  the  special  equili- 
brium polygon,  and  the  lines  of  the  corresponding  force- 
diagram,  lead  directly  to  a  knowledge  of  the  stresses  in  all 
the  cross-sections  of  the  curved  beam  under  consideration, 
bearing  a  given  load ;  or,  vice  versa,  leads  to  a  statement 
of  conditions  to  be  satisfied  by  the  dimensions  of  the  rib, 
for  proper  security. 

It  is  here  supposed  that  the  rib  has  sufficient  la-ceral 


444  MECHANICS   OF   ENGINEERING. 

bracing  (with  others  which  lie  parallel  with  it)  to  prevent 
buckling  sideways  in  any  part  like  a  long  column.  Before 
proceeding  to  the  complete  graphical  analysis  of  the  differ- 
ent cases  of  arch-ribs,  it  will  be  necessary  to  devote  the 
next  few  paragraphs  to  developing  a  few  analytical  rela- 
tions in  the  theory  of  flexure  of  a  curved  beam,  and  to 
giving  some  processes  in  "  graphical  arithmetic." 

369.  Change  in  the  Angle  Between  Two  Consecutive  Rib  Tan- 
gents when  the  rib  is  loaded,  as  compared  with  its  value 
before  loading.  Consider  any  small  portion  (of  an  arch 
rib)  included  between  two  consecutive  cross- sections;  Fig. 
407.  KHG  W  is  its  unstrained  form.  Let  EA,  =  ds,  be 
the  original  length  of  this  portion  of  the  rib  axis.  The 
length  of  all  the  fibres  (j|  to  rib-axis)  was  originally  =ds 
(nearly)  and  the  two  consecutive  tangent-lines,  at  E  and  A, 
made  an  angle  =  dti  originally,  with  each  other.  While 
under  strain,  however,  all  the  fibres  are  shortened  equally 
an  amount  dAlt  by  the  uniformly  distributed  tangential 
thrust,  but  are  unequally  shortened  (or  lengthened,  accord- 
ing as  they  ai  e  on  one  side  or  the  other  of  the  gravity  axis 
E,  or  A,  of  the  section)  by  the  system  of  forces  making 
what  we  call  the  "  stress  couple,"  among  which  the  stress 
at  the  distance  e  from  the  gravity  axis  A  of  the  section  is 
called  pi  per  square  inch ;  so  that  the  tangent  line  at  A' 
now  takes  the  direction  A'D  "\  to  H'A'G'  instead  of  A'Q 
(we  suppose  the  section  at  E  to  remain  fixed,  for  convani- 


ARCH   RIBS.  445 

ence,  since  the  change  of  angle  between  the  two  tangents 
depends  on  the  stresses  acting,  and  not  on  the  new  posi- 
tion in  space,  of  this  part  of  the  rib),  and  hence  the  angle 
between  the  tangent-lines  at  E  and  A  (originally  =  dd]  is 
now  increased  by  an  amount  CA'D  =  dy  (or  G'A'R  =  d(p}\ 
G'H'  is  the  new  position  of  GH.  We  obtain  the  value  of 
d<f>  as  follows:  That  part  (rf/2)  of  the  shortening  of  the 
fibre  at  G,  at  distance  e  from  A  due  to  the  force  pjlF,  is 

§  201  eq.  (1),  dL  =        ?.  But,  geometrically,  d^  also  =ed<p, 


(1.) 


But,  letting  M  denote  the  moment  of  the  stress-couple 
at  section  A  (M  depends  on  the  loading,  mode  of  support, 
etc.,  in  any  particular  case)  we  know  from  §  295  eq.  (6)  that 

M=^-,  and  hence  by  substitution  in  (1)  we  have 


-,       Mds 


/0x 


[If  the  arch-rib  in  question  has  less  than  three  hinges, 
the  equal  shortening  of  the  fibres  due  to  the  thrust  (of 
the  block  in  last  figure)  piF,  will  have  an  indirect  effect  on 
the  angle  dy.  This  will  be  considered  later.] 

370.  Total  Change  |ie./^f  1  in  the  Angle  Between  the  End 

Tangents  of  a  Rib,  before  and  after  loading.      Take  the  ex- 
ample in  Fig.  408  of  a  rib  fixed  at  one  end  and  hinged  at 


446 


MECHANICS   OF    ENGINEERING. 


the  other.  When  the  rib  is  unstrained  (as  it  is  supposed 
to  be,  on  the  left,  its  own  weight  being  neglected ;  it  is  not 
supposed  sprung  into  place,  but  is  entirely  without  strain) 
then  the  angle  between  the  end-tangents  has  some  value 

6'  =    C  cW=  the  sum  of  the  successive  small  angles  dd  for 

e/o 

each  element  ds  of  the  rib  curve  (or  axis).  After  loading, 
[on  the  right,  Fig.  408],  this  angle  has  increased  having 
now  a  value 


'+    r 


i.e.,  a  value  = 


(I.) 


FIG.  409. 


There  must  oe  no  hinge  between  0 


§  371.  Example  of  Equation  (L)  in  Anal- 
ysis. —  A  straight,  homogeneous,  pris- 
matic beam,  Fig.  409,  its  own  weight 
neglected,  is  fixed  obliquely  in  a  wall. 
After  placing  a  load  P  on  the  free  end, 
required  the  angle  between  the  end- 
tangents.  This  was  zero  before  load- 
ing .'.  its  value  after  loading  is 


=o+f'=o+ 


Mds 


By  considering  free  a  portion  between  0  and  any  da  of  the 
beam,  we  find  that  M—  Poj=mom.  of  the  stress  couple. 
The  flexure  is  so  slight  that  the  angle  between  any  ds  and 
its  dx  is  still  practically  =a  (§  364),  and  .-.  ds=dx  sec  a. 
Hence,  by  substitution  in  eq.  (I.)  we  have 


=J_  r°Mds= 

El  Jo 


,=:P(ooBa)P  [Compare  with  §  237]. 


ARCH  RIBS.  447 

It  is  now  apparent  that  if  both  ends  of  an  arch  rib  are 
fixed,  when  unstrained,  and  the  rib  be  then  loaded  (within 
elastic  limit,  and  deformation  slight)  we  must  have 

C*(Md8-*-EI)  =  zero,  since  u'  =0. 

e/o 

372.  Projections  of  the  Displacement  of  any  Point  of  a  Loaded 
Rib  Relatively  to  Another  Point  and  the  Tangent  Line  at  the  Lat- 
ter. —  (There  must  be  no  hinge  between  0  and  B).  Let  0 
be  the  point  whose  displacement  is  considered  and  B  the 
other  point.  Fig.  410.  If  J5's  tangent-line  is  fixed  while 
the  extremity  0  is  not  supported  in  any  way  (Fig.  410) 
then  a  load  P  put  on,  0  is  displaced  to  a  new  position  On. 


With  0  as  an  origin  and  OB  as  the  axis  of  X,  the  projec- 
tion of  the  displacement  00a  upon  X,  will  be  called  J», 
that  upon  Y,  Ay. 

In  the  case  in  Fig.  410,  O's  displacement  with  respect  to 
B  and  its  tangent-line  BT,  is  also  its  absolute  displacement 
in  space,  since  neither  B  nor  BT  has  moved  as  the  rib 
changes  form  under  the  load.  In  Fig.  411,  however,  the 
extremities  0  and  B  are  both  hinged  to  piers,  or  supports, 
the  dotted  line  showing  its  form  when  deformed  under  a 
load.  The  hinges  are  supposed  immovable,  the  rib  being 
free  to  turn  about  them  without  friction.  The  dotted  line 
is  the  changed  form  under  a  load,  and  the  absolute  dis- 
placement of  0  is  zero  ;  but  not  so  its  displacement  rela- 
tively to  B  and  5's  tangent  BT,  for  BT  has  moved  to  a 
new  position  BT'.  To  find  this  relative  displacement  con- 
ceive the  new  curve  of  the  rib  superposed  on  the  old  in 
a  way  that  B  and  BT  may  coincide  with  their  original  po- 


448 


MECHANICS   OF   ENGINEERING. 


sitions,  Fig.  412.  It  is  now  seen  that  O's  displacement 
relatively  to  B  and  BT  is  not  zero  but  =00n,  and  has  a 
small  Ax  but  a  comparatively  large  Jy.  In  fact  for  this 
case  of  hinged  ends,  piers  immovable,  rib  continuous  be- 
tween them,  and  deformation  slight,  we  shall  write  Jo?  = 
zero  as  compared  with  Ay,  the  axis  X passing  through  OB}. 

373.  Values  of  the  X  and  Y  Projections  of  O's  Displacement  Rela- 
tively to  Band  B's  Tangent;  the  origin  being  taken  at  0. 
Fig.  413.  Let  the  co- 
ordinates of  the  dif- 
ferent points  E,  D,  G, 
etc.,  of  the  rib,  re- 
ferred to  0  and  an 
arbitrary  X  axis,  be 
x  and  y,  their  radial 
distances  from  0  be- 
ing u  (i.e.,  u  for  G,  u' 
for  D,  etc.;  in  gener-  | 
al,  u).  OEDG  is  the  i, 
unstrained  form  of  the 
rib,  (e.g.,  the  form  it  Pl'G  413 

would  assume  if  it  lay  flat  on  its  side  on  a  level  platform, 
under  no  straining  forces),  while  OnE"D'CB  is  its  form 
under  some  loading,  i.e.,  under  strain.  (The  superposi- 
tion above  mentioned  (§  372)  is  supposed  already  made  if 
necessary,  so  that  BT  is  tangent  at  B  to  both  forms). 
Now  conceive  the  rib  OB  to  pass  into  its  strained  condi- 
tion by  the  successive  bending  of  each  ds  in  turn.  The 
straining  or  bending  of  the  first  ds,  BG,  through  the  small 
angle  dtp  (dependent  on  the  moment  of  the  stress  couple 
at  G  in  the  strained  condition)  causes  the  whole  finite  piece 
06Yto  turn  about  G  as  a  centre  through  the  same  small 
angle  d<p ;  hence  the  point  0  describes  a  small  linear  arc 
00' —dv,  whose  radius  =  u  the  hypothenuse  of  the  x  and 
y  of  G,  and  whose  value  /.  is  dv=ud</>. 

Next   let  the  section  D,  now  at  D',  turn  through   it» 
proper  angle  d<p'  (dependent  on  its  stress-couple)  carrying 


ARC  II   KIBS.  449 

with  it  the  portion  D'O',  into  the  position  D'0"t  making 
0'  describe  a  linear  arc  0'0"=(dv)'  =u'dy',  in  which  u'  = 
the  hypothenuse  on  the  x'  and  y'  (of  D),  (the  deformation 
is  so  slight  that  the  co-ordinates  of  the  different  points 
referred  to  0  and  X  are  not  appreciably  affected).  Thus, 
each  section  having  been  allowed  to  turn  through  the  an- 
gle proper  to  it,  0  finally  reaches  its  position,  On,  of  dis- 
placement. Each  successive  dv,  or  linear  arc  described  by 
0,  has  a  shorter  radius.  Let  dx,  (dx)',  etc.,  represent  the 
projections  of  the  successive  (dy)'s  upon  the  axis  X\  and 
similarly  %,  (8y)'  etc.,  upon  the  axis  Y.  Then  the  total  X 
projection  of  the  curved  line  0  ....  0,,  will  be 

Jx=     I  dx  and  similarly  Ay  —    /  dy     .     .     .     (1) 

But  d  v  =  u  d  y>,  and  from  similar  right-triangles, 
8  x  :  dv  :  :  y  :  u  and  dy  :  dv  :  :  x  :  u  .'.  dx  =  ydy  and  dy=xd<p  ; 
whence,  (see  (1)  and  (2)  of  §369) 


If  the  rib  is  homogeneous  E  is  constant,  and  if  it  is  of 
constant  cross-section,  all  sections  being  similarly  cut  by 
the  vertical  plane  of  the  rib's  axis  (i.e.,  if  it  is  a  "  curved 
prism  ")  7,  the  moment  of  inertia  is  also  constant. 

374.  Recapitulation  of  Analytical  Relations,   for    reference* 
(Not  applicable  if  there  is  a  hinge  between  0  and  E) 

Total  Change  in  Angle  between  )  _   f**Mds  fT  . 

tangent-lines    0   and   B      \  ~J0~EI 

The  X-Projection  of  O's  Displacement  "] 

Relatively  to  B.and  Fs  tangent-  »Md8 

line  ;    (the  origin  being  at    0)  L  —  I       *•  •     • 
and  the  axes  X  and  Y  ~\  to  f 
each  other) 


450 


MECHANICS  OF   ENGINEERING. 


The  Y-Projection  of  O's  Displacement,  )  _ 
etc.,  as  above. 


(III.) 


Here  x  and  y  are  the  co-ordinates  of  points  in  the  rib- 
curve,  ds  an  element  of  that  curve,  M,  the  moment  of  the 
stress-couple  in  the  corresponding  section  as  induced  by 
the  loading,  or  constraint,  of  the  rib. 

(The  results  already  derived  for  deflections,  slopes,  etc., 
for  straight  beams,  could  also  be  obtained  from  these 
formulae,  L,  II.  and  III.  In  these  formulae  also  it  must 
be  remembered  that  no  account  has  been  taken  of  the 
shortening  of  the  rib-axis  by  the  thrust,  nor  of  the  effect 
of  a  change  of  temperature.) 

374a.  Resumfe  of  the  Properties  of  Equilibrium  Polygons  and 
their  Force  Diagrams,  for  Systems  of  Vertical  Loads. — See  §§  335 
to  343.  Given  a  system  of  loads  or  vertical  forces,  Plt  P2, 

etc.,    Fig.   414,    and 
P,  two  abutment  verti- 
cals, F'  and   G' ;    if 
we  lay  off,  vertically, 
to    form    a    "  load- 
line,"  1 . .  2  =  Plt  2. . . 
P3  3=P2>etc.,  select  any 


Pole,  0,,  and  join  0j 
...  1,   0i  ...  2,  etc. ; 
also,    beginning     at 
FIG-  414-  any  point  FI  in  the 

vertical  F',  if  we  draw  Fl . . .  a  \\  to  Ol . .  1  to  intersect  the 
line  of  Pl ;  then  ab  \\  to  0, .  .  2,  and  so  on  until  finally  a 
point  GI,  in  G',  is  determined;  then  the  figure  FI  .abc  Cnis 
an  equilibrium  polygon  for  the  given  loads  and  load  verti- 
cals, and  0j ...  1234  is  its  "  force  diagram."  The  former 
is  so  called  because  the  short  segments  F&  ab,  etc.,  if 
considered  to  be  rigid  and  imponderable  rods,  in  a  vertical 
plane,  hinged  to  each  other  and  the  terminal  ones  to  abut- 
ments F{  and  #„  would  be  in  equilibrium  under  the  given 
loads  hung  at  the  joints.  An  infinite  number  of  equilib- 


AKCH-RIBS.  451 

riura  polygons  may  be  drawn  for  the  given  loads  and 
abutment-verticals,  by  choosing  different  poles  in  the  force 
diagram.  [One  other  is  shown  in  the  figure  ;  02  is  its 
pole.  (J^  G!  and  F2  G2  are  abutment  lines.)]  For  all  of 
these  the  following  statements  are  true  : 

(1.)  A  line  through  the  pole,  ||  to  the  abutment  line  cuts 
the  load-line  in  the  same  point  n't  whichever  equilibrium 
polygon  be  used  (  /.  any  one  will  serve  to  determine  n'). 

(2.)  If  a  vertical  CD  be  drawn,  giving  an  intercept  z'  in 
each  of  the  equilibrium  polygons,  the  product  Hz'  is  the 
same  for  all  the  equilibrium  polygons.  That  is,  (see  Fig. 
414)  for  any  two  of  the  polygons  we  have 

H{  :#2::s/:  V;  or  H2z2'  =Hlzlf. 

(3.)  The  compression  in  each  rod  is  given  by  that 
"  ray  "  (in  the  force  diagram)  to  which  it  is  parallel. 

(4.)  The  "  pole  distance  "  H,  or  ~|  let  fall  from  the  pole 
upon  the  load-line,  divides  it  into  two  parts  which  are  the 
vertical  components  of  the  compressions  in  the  abutment- 
rods  respectively  ( the  other  component  being  horizontal) ; 
//  is  the  horizontal  component  of  each  (and,  in  fact,  of 
each  of  the  compressions  in  all  the  other  rods).  The 
compressions  in  the  extreme  rods  may  also  be  called  the 
abutment  reactions  (oblique)  and  are  given  by  the  extreme 
rays. 

(5.)  Three  Points  [not  all  in  the  same  segment  (or  rod)] 
determine  an  equilibrium  polygon  for  given  loads.  Hav- 
ing given,  then,  three  points,  we  may  draw  the  equilibrium 
polygon  by  §341. 

375.  Summation  of  Products.  Before  proceeding  to  treat 
graphically  any  case  of  arch-ribs,  a  few  processes  in 
graphical  arithmetic,  as  it  may  be  called,  must  be  pre- 
sented, and  thus  established  for  future  use. 

To  make  a  summation  of  products  of  two  factors  in  each 
by  means  of  an  equilibrium  polygon. 


452 


MECHANICS   OF   ENGINEERING. 


Construction.     Suppose  it  required  to  make  the  summa- 
tion 21  (x  z)  i.  e.y  to  sum  the  series 

xl  Zj-f-  x.z  z2  +  a?3  %  +    .     .     .     by  graphics. 

Having  first  arranged  the  terms  in  the  order  of  magni- 
tude of  the  a?'s,  we  proceed  as  follows :  Supposing,  for 
illustration,  that  two  of  the  z'a  (%  and  z4)  are  negative 
(dotted  in  figure)  see  Fig.  415.  These  quantities  x  and  z 
may  be  of  any  nature  whatever,  anything  capable  of  being 
represented  by  a  length,  laid  off  to  scale. 

First,  in  Fig. 
416,  lay  off  the 
z's  in  their 
order,  end  to 
end,  on  a  ver- 
tical load-line 
taking  care  to 
J  off  z3  and 

i J_  24   upicard    in 

£«_their   turn. 
Take  any  con- 
FIG.  416.  venient     pole 

1,  0  ...  2,  etc.;  then,  having  pre- 
viously drawn  vertical  lines  whose  horizontal  distances 
from  an  extreme  left-hand  vertical  F'  are  made  =  x{)  x,r 
a?3,  etc.,  respectively,  we  begin  at  any  point  F,  in  the  verti- 
cal F't  and  draw  a  line  ||  to  0  ...  1  to  intersect  the  x}  ver- 
tical in  some  point  ;  then  1'  2'  II  to  0  ...  2,  and  so  on,  fol- 
lowing carefully  the  proper  order.  Produce  the  last  seg- 
ment (6'  ...  G  in  this  case)  to  intersect  the  vertical  F'  in 
some  point  K.  Let  KF  =k  (measured  on  the  same  scale 
as  the  x's),  then  the  summation  required  is 

S«  (xz)  =  Hk. 

H  is  measured  on  the  scale  of  the  z's,  which  need  not  be 
the  same  as  that  of  the  a?'s  ;  in  fact  the  z's  may  not  be  the 
same  kind  of  quantity  as  the  o?'s. 

[PROOF. — From  similar  triangles  H:  2, ::  xl :  &„  .'.  x^^Hlc,  \ 
and  "  "  "         H:z*::x2:  k.2,  .-.  xjs.2=Hk, , 


0 ;  draw  the  rays  0  . 


-CH-RIBS. 


453 


and  so  on.     But  H(kl-\-k,+etG.}=HxFK=Hk]. 

376.  Gravity  Vertical — From  the  same  construction  in 
Fig.  415  we  can  determine  the  line  of  action  (or  gravity 
vertical)  of  the  resultant  of  the  parallel  vertical  forces  zlt 
%,  etc.-  (or  loads);  by  prolonging  the  first  and  last  segments 

to  their  intersection  at 
(j.  The  resultant  of  the 
system  of  forces  or  loads 
acts  through  C  and  is 
vertical  in  this  case  ;  its 
value  .being  =  JT  (2), 
that  is,  it  =  the  length 

1  ...  7  in  the  force  dia- 
gram, interpreted  by  the 
proper  scale.     It  is  now 
supposed    that    the   z's 
represent  forces,  the  a?'s 
being    their    respective 
lever  arms  about  F.     If 
the     z's    represent     the 
areas  of  small  finite  por- 
tions  of   a   large  plane 
figure,   we   may   find    a 
gravity -line  (through  (7) 
of   that    figure    by    the 
above  construction;  each 

2  being-applied  through 
the  centre  of  gravity  of 
its  own  portion. 

Calling  the    distance 
a?  between  the  verticals 
through    C  jmd    F,  we 
t,  have    also   x  .  2  (2)    = 
I  (xz)  because   I  (z)  is 
the  resultant  of  the  II  z's. 
This  is  also  evident  from 
the  proportion  (similar 
triangles) 
H  :  (I  .  .  7)  ::  x  :  L 


454  MECHANICS   OF   ENGINEERING. 

376  a.  Moment  of  Inertia  (of  Plane  Figure)  by  Graphics.— Fig. 
416  a.  Iy  =  ?  First,  for  the  portion  on  right.  Divide  OR 
jnto  equal  parts  each  =  Ax.  Let  z},  z.2,  etc.,  be  the  middle 
ordinates  of  the  strips  thus  obtained,  and  xl}  etc.  their 
abscissas  (of  middle  points). 

Then  we  have  approximately 


/N  for  OH  =  AX.Z&I+  Ax.z.2x.?+  ...... 

•••]..  (1) 


But  by  §375  we  may  construct  the  products  z^x^z^x^  etc., 
taking  a  convenient  H',  (see  Fig.  416,  (6)),  and  obtain  Jcl}  Jc2> 
etc.,  such  that  ZjXi  =  H'klt  z?x2  =  H'k2,  etc.  Hence  eq.  (1) 
becomes  : 

7,  for  OB  approx.=7Pzte[&1;r1+&2a?2+  ...]...  (2) 

By  a  second  use  of  §  375  (see  Fig.  416  c)  we  construct  I, 
such  that  lew  +  k.2x2  +  ____  =  H"l  \H"  taken  at  con- 
venience]. /.  from  eq.  (2)  we  have  finally,  (approx.), 

7N  for  OR=H'H"lAx  ____  (3) 

For  example  if  OR  ~  4  in.,  with  four  strips,  Ax  would  = 
1  in.;  and  if  H'  =  2  in.,  H"  =  2  in.,  and  I  =  5.2  in.,  then 

7N  for  OR  =  2x2x5.2x1.0=20.8  biquad.  inches. 

The  7N  for  OL,  on  the  left  of  N,  is  found  in  a  similar 
manner  and  added  to  7X  for  OR  to  obtain  the  total  7V  The 
position  of  a  gravity  axis  is  easily  found  by  cutting  the 
shape  out  of  sheet  metal  and  balancing  on  a  knife  edge  ;  or 
may  be  obtained  graphically  by  §  336  ;  or  376. 

377.  Construction  for  locating  a  line  vm(Fig.  417)  at  (a),  in 
the  polygon  FG  in  such  a  position  as  to  satisfy  the  two 
following  conditions  with  reference  to  the  vertical  inter- 
cepts at  1,  2,  3,  4,  and  5,  between  it  and  the  given  points 
1,  2,  3,  etc.,  of  the  perimeter  of  the  polygon. 


ARCH-RIBS. 


455 


Condition  I — (Calling  these  intercepts  w,,  u^,  etc.,  and  their 
horizontal  distances  from  a  given  vertical  F,  xlt  x.,,  etc.) 

2  (u)  is  to  =  0;  i.e.,  the  sum  of  the  positive  w's  must  be 
numerically  =  that  of  the  negative  (which  here  are  at  1 
and  5).  An  infinite  number  of  positions  of  vm  will  satisfy 
condition  I. 


Condition  H — S  (ux)  is  to  =  0  ;  i.e.,  the  sum  of  the 
moments  of  the  positive  w's 
about  F  must  =  that  of  the 
negative  w's.  i.e.,  the  moment 
of  the  resultant  of  the  posi- 
tive w's  must  =  that  of  the 
resultant  of  the  negative ; 
and  .'.  (Condit.  I  being 
already  satisfied)  these  two 
resultants  must  be  directly 
opposed  and  equal.  But  the 
ordinates  u  in  (a)  are  indi- 
vidually equal  to  the  differ- 
ence of  the  full  and  dotted 
ordinates  in  (b)  with  the 
same  x's  .'.  the  conditions 
may  be  rewritten  : 

I.  2'  (full  ords.  in(Z>))= 
I  (dotted  ords.  in  (b)) 

II.  2  [each  full  ord.  in  (6) 
X  its  x]   =  -    [each  dotted 
ord.  in  (b)   x  its  x]  i.e.,  the 
centres  of  gravity  of  the  full 

and  of  the  dotted  in  (b)  must  lie  in  the  same  vertical 

Again,  by  joining  vG,  we  may  divide  the  dotted  ordi- 
nates of  (b)  into  two  sets  which  are  dotted,  and  broken,  re- 
spectively, in  (c)  Then,  finally,  drawing  in  (cZ), 

R,  the  resultant  of  full      ords.  of  (c) 
T,    "  "          "  broken   "       "   " 

T',  "          "          "  dotted    "      "  " 


^^j 

|         ! 

T      ! 

R                    f, 

i  1 

T 

('0 


FIG.  417. 


456  MECHANICS   OF   ENGINEERING. 

we  are  prepared  to  state  in  still  another  and  final  form  the 
conditions  which  vm  must  fulfil,  viz. : 

(I.)  T+T'  must  =  R;  and  (II.)  The  resultant  of  T 
and  T'  must  act  in  the  same  vertical  as  R. 

In  short,  the  quantities  T,  T',  and  R  must  form  a  bal- 
anced system,  considered  as  forces.  All  of  which  amounts 
practically  to  this :  that  if  the  verticals  in  which  T  and  T' 
act  are  known  and  R  be  conceived  as  a  load  supported  by 
a  horizontal  beam  (see  foot  of  Fig.  417,  last  figure)  resting 
on  piers  in  those  verticals,  then  T  and  T'  are  the  respec- 
tive reactions  of  those  piers.  It  will  now  be  shown  that  the 
verticals  of  T  and  T'  are  easily  found,  being  independent  of 
the  position  of  vm;  and  that  both  the  vertical  and  the  mag- 
nitude of  R,  being  likewise  independent  of  vm,  are  deter- 
mined with  facility  in  advance.  For,  if  v  be  shifted  up 
or  down,  all  the  broken  ordinates  in  (c)  or  (d)  will  change 
in  the  same  proportion  (viz.  as  vF  changes),  while  the 
dotted  ordinates,  though  shifted  along  their  verticals,  do 
not  change  in  value  ;  hence  the  shifting  of  v  affects  neither 
the  vertical  nor  the  value  of  T',  nor  the  vertical  of  T. 
The  value  of  T,  however,  is  proportional  to  vF.  Similar- 
ly, if  in  be  shifted,  up  or  down,  T'  will  vary  proportionally 
to  mG,  but  its  vertical,  or  line  of  action,  remains  the  same. 
T  is  unaffected  in  any  way  by  the  shifting  of  m.  R,  de- 
pending for  its  value  and  position  on  the  full  ordinates  of 
(c)  Fig.  417,  is  independent  of  the  location  of  vm.  We 
may  /.  proceed  as  follows  : 

1st.  Determine  R  graphically,  in  amount  and  position, 
by  means  of  §  376. 

2ndly.  Determine  the  verticals  of  T  and  T'  by  any  trial 
position  of  vm  (call  it  v2w2),  and  the  corresponding  trial 
values  of  T  and  T  (call  them  T2  and  T'2). 

3rdly.  By  the  fiction  of  the  horizontal  beam,  construct 
(§  329)  or  compute  the  true  values  of  T  and  T',  and  then 
determine  the  true  distances  vF  and  mG  by  the  propor- 
tions 

vF  :  v,F  :  :  T  :  T,  and  mG  :  m»G  :  :  T'  :  T'z. 


ARCH-RIBS. 


457 


Example  of  this.     Fig.  418.      (See  Fig.  417  for  s  and  t.) 

From  A  toward  B  in  (e)  Fig.  418,  lay  off  the  lengths  (or 
lines  proportional  \* 

to  them)  of  the  full  °/f\ 

ordinates  1,  2,  etc., 
of  (/).  Take  any 
pole  Oi,  and  draw  the 
equilibrium  p  o  1  y  - 
g  o  n  (/")'  and  pro- 
long its  extreme  seg- 
ments to  find  C  and 
thus  determine  7?'s 
vertical.  R  is  repre- 
sented by  AB.  In 
(g)  [same  as  (/)  but 
shifted  to  avoid 
complexity  of  lines] 
draw  a  trial  v.,m.,  and 
join  v2  G.,.  Deter- 
mine the  sum  T2  of 
the  broken  ordi-  'FIG.  4is. 

nates  (between  v.2G2  ana  F2G2)  and  its  vertical  line  of  ap- 
plication, precisely  as  in  dealing  with  R ;  also  T'2  that  of 
the  dotted  ordiuates  (five)  and  its  vertical.  Now  the  true 
T=Rt+(8+t)  and  the  true  T'=Rs+(s+t).  Hence  com- 
pute vF=(T+T2}  v,F2  and  mG=(T'+T't}  m^G2,  and  by 
laying  them  off  vertically  upward  from  F  and  G  respec- 
tively we  determine  v  and  m,  i.e.,  the  line  vm  to  fulfil  the 
conditions  imposed  at  the  beginning  of  this  article,  rela- 
ting to  the  vertical  ordinates  intercepted  between  vm  and 
given  points  on  the  perimeter  of  a  polygon  or  curve. 

Note  (aX  If  the  verticals  in  which  the  intercepts  lie  are 
equidistant  and  quite  numerous,  then  the  lines  of  action 
of  T2  and  T'2  will  divide  the  horizontal  distance  between 
F  and  G  into  three  equal  parts.  This  will  be  exactly  true 
in  the  application  of  this  construction  to  §  390. 

Note  (b).  Also,  if  the  verticals  are  symmetrically  placed 
about  a  vertical  line,  (as  will  usually  be  the  case)  tww2  is 


458  MECHANICS    OF    ENGINEERING. 

best  drawn  parallel  to  FG,  for  then  T^  and  T'3  will  be 
equal  and  equi-distant  from  said  vertical  line. 

378.  Classification  of  Arch-Bibs,  or  Elastic  Arches,  according 
to  continuity  and  modes  of  support.  In  the  accompany- 
ing figures  ihefull  curves  show  the  unstrained  form  of  the 
rib  (before  any  load,  even  its  own  weight,  is  permitted  to 
come  upon  it)  ;the  dotted  curve  shows  its  shape  (much  ex- 
aggerated) when  bearing  a  load.  For  a  given  loading 
Three  Conditions  must  be  given  to  determine  the  special 
equilibrium  polygon  (§§  366  and  367). 

Class  A. — Continuous  rib,  free  to  slip  laterally  on  the 
piers,  which  have  smooth  horizontal  surfaces,  Fig.  420. 

This  is  chiefly  of  theoretic  interest,  its  consideration 
being  therefore  omitted.  The  pier  reactions  are  neces- 
sarily vertical,  just  as  if  it  were  a  straight  horizontal 
beam. 

Class  B.  Rib  of  Three  Hinges,  two  at  the  piers  and  one 
intermediate  (usually  at  the  crown)  Fig.  421.  Fig.  36  also 
is  an  example  of  this.  That  is,  the  rib  is  discontinuous 
and  of  two  segments.  Since  at  each  hinge  the  moment  of 
the  stress  couple  must  be  be  zero,  the  special  equilibrium 
polygon  must  pass  through  the  hinges.  Hence  as  three 
points  fully  determine  an  equilibrium  polygon  for  given 
load,  the  special  equilibrium  is  drawn  by  §  341. 


FIG.  420.  FIG.  421. 

[§  378a  will  contain  a  construction  for  arch-ribs  of  three 
hinges,  when  the  forces  are  not  all  vertical.] 

Class  C.  Rib  of  Two  Hinges,  these  being  at  the  piers,  the 
rib  continuous  between.  The  piers  are  considered  im- 
movable, i.e.,  the  span  cannot  change  as  a  consequence  of 
loading.  It  is  also  considered  that  the  rib  is  fitted  to  its 


AllCil   KIBS.  459 

hinges  at  a  definite  temperature,  and  is  then  under  no  con- 
straint from  the  piers  (as  if  it  lay  flat  on  the  ground),  not 
even  its  own  weight  being  permitted  to  act  when  it  is  fi- 
nally put  into  position.  AVhen  the  "false  works" 
or  temporary  supports  are  removed,  stresses  are  in- 
duced in  the  rib  both  by  its  loading,  including  its 
own  weight,  and  by  a  change  of  temperature.  Stresses 
due  to  temperature  may  be  ascertained  separately  and 
then  combined  with  those  duo  to  the  loading.  [Classes 
A  and  B  are  not  subject  to  temperature  stresses.]  Fig. 

422  shows  a  rib  of  two  hinges, 
at  ends.  Conceive  the  dotted 
curve  (form  and  position  un- 
der strain)  to  be  superposed 
on  the  continuous  curve 
(form  before  strain)  in  such 
a  way  that  B  and  its  tangent 
FIG-  422-  line  (which  has  been  dis- 

placed from  its  original  position)  may  occupy  their  pre- 
vious position.  This  gives  us  the  broken  curve  OnB.  00a 
is  .'.  O's  displacement  relatively  to  B  and  .Z?'s  tangent. 
Now  the  piers  being  immovable  OnB  (right  line)=O.B  ;  i.e., 
the  X  projection  (or  Ax)  of  00,,  upon  OB  (taken  as  an  axis 
of  X}  is  zero  compared  with  its  Ay.  Hence  as  one  condi- 
tion to  fix  the  special  equilibrium  polygon  for  a  given  load- 
ing we  have  (from  §  373) 


r 


(1) 


The  other  two  are  that  the  )  must  pass  through  0  .  (2) 
special  equilibrium  polygon  )  "  "  "  B  .  (3) 

Class  D.  Rib  with  Fixed  Ends  and  no  hinges,  i.e.,  continu- 
ous. Piers  immovable.  The  ends  may  be  fixed  by  being 
inserted,  or  built,  in  the  masonry,  or  by  being  fastened  to 
large  plates  which  are  bolted  to  the  piers.  [The  St.  Louis 
Bridge  and  that  at  Coblenz  over  the  Rhine  are  of  this 
class.]  Fig.  423.  In  this  class  there  being  no  hinges  we 


400 


MECHANICS   OF   ENGLNEE11IXG. 


have  no  point  given  in  advance  through  which  the  special 
equilibrium  polygon  must  pass.  However,  since  O's  dis- 
placement relatively  (and  absolutely)  to  B  and  J?'s  tangent 
is  zero,  both  Ax  and  Ay  [see  §  373]  —  zero.  Also  the  tan- 
gent-lines both  at  0  and  B  being 
fixed  in  direction,  the  angle  be- 
tween them  is  the  same  under 
loading,  or  change  of  temperature, 
as  when  the  rib  was  first  placed 
in  position  under  no  strain  and  at 
a  definite  temperature. 
Hence  the  conditions  for  locating  the  special  equilibrium 
polygon  are 


FIG.  423. 


Mds  _ 
El 


ls  =  o  •    r  Myas  —  o  •  c 
r  "       c/o  #/        '  Jo 


Mxds 


=  0. 


£7  e/o      El 

In  the  figure  the  imaginary  rigid  prolongations  at   the 
ends  are  shown  [see  §  366]. 

Other  designs  than  those  mentioned  are  practicable 
(such  as :  one  end  fixed,  the  other  hinged ;  both  ends  fixed 
and  one  hinge  between,  etc.),  but  are  of  unusual  occur- 
rence. 

378a.  Rib  of  Three  Hinges.  Forces  not  all  Vertical.  If  the 
given  rib  of  three  hinges  upholds  a  roof,  the  wind-press- 
ure on  which  is  to  be  considered  as  well  as  the  weights  of 
the  materials  composing  the  roof-covering,  the  forces  will 
not  all  be  vertical.  To  draw  the  special  equil.  polygon  in 

such  a  case  the  following 
construction  holds  :  Re- 
quired to  draw  an  equilib- 
rium polygon,  for  any 
plane  system  of  forces, 
through  three  arbitrary 
points,  A,  p  and  B ;  Fig. 
B423a.  Find  the  line  of 
action  of  7?,,  the  resultant 
of  all  the  forces  occurring 
FIG.  423a.  ^  between  A  and  p ;  also, 


AKCH-RLBS.  401 

that  of  R2,  the  resultant  of  all  forces  between  ,p  and  B  ; 
also  the  line  of  action  of  R,  the  resultant  of  BI  and  R2,  [see 
§  328.]  Join  any  point  M  in  R  with  A  and  also  with  B, 
and  join  the  intersections  N&nd  0.  Then  A  N  will  be  the 
direction  of  the  first  segment,  0  B  that  of  the  last,  and 
NO  itself  is  the  segment  corresponding  to  p  (in  the  de- 
sired polygon)  of  an  equilibrium  polygon  for  the  given 
forces.  See  §  328.  If  A  N' p  0'  B  are  the  corresponding 
segments  (as  yet  unknown)  of  the  desired  equil.  polygon, 
we  note  that  the  two  triangles  MNO  and  M'N'C/,  having 
their  vertices  on  three  lines  which  meet  in  a  point  [i.e.,  R 
meets  Rl  and  R2  in  C"],  are  homological  [see  Prop.  VII.  of 
Introduc.  to  Modern  Geometry,  in  Chauvenet's  Geometry,] 
and  that  .  • .  the  three  intersections  of  their  corresponding 
sides  must  lie  on  the  same  straight  line.  Of  those  inter  ^ 
sections  we  already  have  A  and  B,  while  the  third  must  be 
at  (7,  found  at  the  intersection  of  AB  and  NO.  Hence  by 
connecting  C  and  p,  we  determine  .A7'  and  0'.  Joining 
WA  and  O'B,  the  first  ray  of  the  required  force  diagram  will 
be  ||  to  NA,  while  the  last  ray  will  be  II  to  O'B,  and  thus 
the  pole  of  that  diagram  can  easily  be  found  and  the  cor- 
responding equilibrium  polygon,  beginning  at  A,  will  pass 
through  p  and  B. 

(This  general  case  includes  those  of  §§  341  and  342.) 
379.  Arch-Rib  of  two  Hinges;  by  Prof.  Eddy's  Method.* 
[It  is  understood  that  the  hinges  are  at  the  ends.]  Re- 
quired the  location  of  the  special  equilibrium  polygon.  We 
here  suppose  the  rib  homogeneous  (i.e.,  the  modulus  of 
Elasticity  E  is  the  same  throughout),  that  it  is  a  "  curved 
prism  "  (i.e.,  that  the  moment  of  inertia  I  of  the  cross- 
section  is  constant),  that  the  piers  are  on  a  level,  and  that 
the  rib-curve  is  symmetrical  about  a  vertical  line.  Fig. 
424.  For  each  point  m  of  the  rib 
curve  we  have  an  x  and  y  (both 
known,  being  the  co-ordinates  of 
the  point),  and  also  a  z  (intercept 
;-f  between  rib  and  special  equilib- 


rium polygon)  and  a  z'  (intercept 


*  P.  25  of  Prof.  Eddy's  book  ;  see  reference  in  preface  of  this  work. 


4G2 


MECHANICS   OF   ENGINEERING. 


between  the  spec.  eq.  pol.  and  the  axis  X  (which  is  OS}. 
The  first  condition  given  in  §  378  for  Class  C  may  be 
transformed  as  follows,  remembering  [§  367  eq.  (3)]  that 
M  =  Hz  at  any  point  m  of  the  rib  (and  that  El  is  con- 
stant). 


CJJL 


Myds  =  0,  i.e.,  *    fzyds  =  0  .  -  .  fzyds  =  0 

JLJ.   e/o  e/o 


but 

z  =  y  —  z' 


'•  C  (y  —  z')yds=0; i.e.,  f  yyds  =  C  yz'ds  .  (1) 

e/o  e/o  e/o 

In  practical  graphics  we  can  not  deal  with  infinitesimals ; 
hence  we  must  substitute  As  a  small  finite  portion  of  the 
rib-curve  for  ds;  eq.  (1)  now  reads  I*  yy  As  =  J?0B  yz'  As. 
But  if  we  take  all  the  As's  equal,  As  is  a  common  factor 
and  cancels  out,  leaving  as  a  final  form  for  eq.  (1) 

S»(yy}  =  Z*(yz'}     .     .        .     (1)' 

The  other  two  conditions  are  that  the  special  equilibrium 
polygon  begins  at  0  and  ends  at  B.  (The  subdivision  of 
the  rib-curve  into  an  even  number  of  equal  As's  will  be  ob- 
served in  all  problems  henceforth.) 

379a.  Detail  of  the  Construction.     Given  the  arch-rib  0  B, 
Fig.  425,  with  specified  loading.     Divide  the  curve  into 


FIG.   425. 


AKCH  RIBS. 


463 


eight  equal  /te's  and  draw  a  vertical  through  the  middle 
of  each.  Let  the  loads  borne  by  the  respective  Js's  be 
PI,  P2,  etc.,  and  with  them  form  a  vertical  load-line  A  C  to 
some  convenient  scale.  With  any  convenient  pole  0" 
draw  a  trial  force  diagram  0"  AC,  and  a  corresponding 
trial  equilibrium  polygon  F  G,  beginning  at  any  point  in 
the  vertical  F.  Its  ordinates  z/',  z2",  etc.,  are  propor- 
tional to  those  of  the  special  equil.  pol.  sought  (whose 
abutment  line  is  OS)  [§374a  (2)].  We  next  use  it  to  de- 
termine n'  [see  §  374a].  We  know  that  OB  is  the  "  abut- 
ment-line "  of  the  required  special  polygon,  and  that  .  •  . 
its  pole  must  lie  on  a  horizontal  through  n'.  It  remains 
to  determine  its  H,  or  pole  distance,  by  equation  (1)'  just 
given,  viz.  :  S*  yy  =  -\yz'.  First  by  §  375  find  the  value 
of  the  summation  -*(yy\  which,  from  symmetry,  we  may 
write  =  2^V 

yl  Hence,  Fig.  426,  we  obtain 


Next,  also  by  §  375,  see  Fig. 
427,  using  the  same  pole  dis- 
tance H0  as  in  Fig.  426,  we 
find 


"  +y* 


Z("  Again,  since  2|    (yz")=  y^" 
+  2/7*7"  +  y<&"  +  2/5*5"  which 
"  from    symmetry  (of  rib) 


we  obtain,  Fig.  428, 
,;  I!(yO  =  #a*/',(Bame    J£); 
and  .-. 

2{  (yz"}=H0  (£,"+&/')•     If  now  we  find  that  kS'+kT"=Zk, 


464  MECHANICS  OF   ENGINEERING. 

the  condition  21  (yy)  =  If  (yz")  is  satisfied,  and  the  pole 
distance  of  our  trial  polygon  in  Fig.  425,  is  also  that  of 
the  special  polygon  sought;  i.e.,  the  z"  's.are  identical  in 
value  with  the  s"s  of  Fig.  424.  In  general,  of  course,  we 
do  not  find  that  &,"+V  —  2&.  Hence  the  z"  's  must  all 
be  increased  in  the  ratio  2&:  (V+V)  to  become  equal  to 
the  z"s.  That  is,  the  pole  distance  H  of  the  spec,  equil- 
polygon  must  be 

TT_  &/'+&/'  rrtr  (in  which  H"  =  the  pole  distance  of  the 
%k  trial  polygon)  since  from  §339  the   ordi- 

nates  of  two  equilibrium  polygons  (for  the  same  loads) 
are  inversely  as  their  pole  distances.  Having  thus  found 
the  Hoi  the  special  polygon,  knowing  that  the  pole  must 
lie  on  the  horizontal  through  n',  Fig.  425,  it  is  easily 
drawn,  beginning  at  0.  As  a  check,  it  should  pass  through 
B. 

For  its  utility  see  §  367,  but  it  is  to  be  remembered  that 
the  stresses  as  thus  found  in  the  different  parts  of  the 
rib  under  a  given  loading,  must  afterwards  be  combined 
with  those  resulting  from  change  of  temperature  and  the 
shortening  of  the  rib  axis  due  to  the  tangential  thrusts, 
before  the  actual  stress  can  be  declared  in  any  part. 

[NoTE. — If  the  "  moment  of  inertia,"  /,  of  the  rib-sec- 
tion is  different  at  different  sections,  i.e.,  if  /is  variable, 

foreq.  (1)'  we  may  write  2*((y.^}= -*(y.  *'}    .     .     .     (1)' 
\      "/  \    -H/ 

(where  n  =  -,  70  being  the  moment  of  inertia  of  a  particu- 

•*0 

lar  section  taken  as  a  standard  and  /that  at  any  section 
of  rib)  and  in  Fig.  426,  use  the  ^  of  each  Js  instead  of  y 

in   the  vertical  "  load-line,"  and  —   for  z"  in  Figs.  427  and 

n 

428].     ' 


ARCH-IUBS. 


465 


330.  Arch  Rib  of  Fixed  Ends  and  no  Hinges,— Example  of 
Class  D.  Prof.  Eddy's  Method.*  As  before,  E  and  I  are 
constant  along  the  rib  Piers  immovable.  Bib  curve 
symmetrical  about  a  vertical  line.  Fig.  429  shows  such  a 
rib  under  any  loading.  Its  span  is  OB,  which  is  taken  as 
an  axis  X.  The  co-ordinates  of  any  point  m'  of  the  rib 
curve  are  x  and  y,  and  z  is  the  vertical  intercept  between 
m'  and  the  special  equilibrium  polygon  (as  yet  unknown, 
but  to  be  constructed).  Prof.  Eddy's  method  will  now  be 

given  for  finding  tha  spe- 
v»    cial  equil.  polygon.     The 
three      conditions     it 
must   satisfy   (see  §  378, 
Class     D,     remembering 
that  E  and  I  are  constant 
and   that   M  —   Ih  fr  jin 
FIG.  429.  §  367)  are 

flds=  0  ;  Cxzds=  ;  0  and    fyzds=0     .     .      (1) 

e/o  e/o  e/o 

Now  suppose  the  auxiliary  reference  line  (straight)  vm 
to  have  been  drawn  satisfying  the  requirements,  with 
respect  to  the  rib  curve  that 

/»B  fP 

/*'ds=0;and      xz'ds=0    ....      (2) 

e/o  e/o 

in  which  z'  is  the  vertical  distance  of  any  point  m'  from 
vm  and  x  the  abscissa  of  m'  from  0. 

From  Fig.  429,  letting  z"  denote  the  vertical  intercept 
(corresponding  to  any  m')  between  the  spec,  polygon  and 
the  auxiliary  line  vm,  we  have  z=zf—z",  hence  the  three 
conditions  in  (1.)  become 


-z"}ds=0',  i.e.,  see  eqs.  (2)          z"ds=Q  . 


(3) 


*  P.  14  of  Prof.  Eddv's  book ;  see  reference  in  preface  of  this  work. 


466  MECHANICS   OF   ENGINEERING. 

Cx  (z'-z")ds=0  ;  i.e.,  see  eqs.  (2)    C*xz"ds=o 

e/o  e/  o 


provided  vm  has  been  located  as  prescribed. 

For  graphical  purposes,  having  subdivided  the  rib  curve 
into  an  even  number  of  small  equal  J,s's,  and  drawn  a  verti- 
cal through  the  middle  of  each,  we  first,  by  §  377,  locate 
vm  to  satisfy  the  conditions 

1?(«0=0  and  I?(a*')=0         .    •    .  •       (6) 

(see  eq.  (2)  ;  the  As  cancels  out)  ;  and  then  locate  the 
special  equilibrium  polygon,  with  vm  as  a  reference-line, 
by  making  it  satisfy  the  conditions. 


2ft«")=0  •  (7);    2Ka*")=0  .  (8);   S*0(yz")=S*0(yz'}  .   (9) 

(obtained  from  eqs.  (3),  (4),  (5)  by  putting  ds=4s,  and  can- 
celling). 

Conditions  (7)  and  (8)  may  be  satisfied  by  an  infinite 
number  of  polygons  drawn  to  the  given  loading.  Any  one 
of  these  being  drawn,  as  a  trial  polygon,  we  determine  for  it 
the  value  of  the  sum  2*(yz")  by  §  375,  and  compare  it  with 
the  value  of  the  sum  2*(yz')  which  is  independent  of  the 
special  polygon  and  is  obtained  by  §  375.  [N.B.  It  must 
be  understood  that  the  quantities  (lengths)  x,  y,  z,  z',  and  z", 
here  dealt  with  are  those  pertaining  to  the  verticals  drawn 
through  the  middles  of  the  respective  ^s's,  which  must  be 
sufficiently  numerous  to  obtain  a  close  result,  and  not  to 
the  verticals  in  which  the  loads  act,  necessarily,  since  these 
latter  may  be  few  or  many  according  to  circumstances,  see 
Fig.  429].  If  these  sums  are  not  equal,  the  pole  distance 
of  the  trial  equil.  polygon  must  be  altered  in  the  proper 
ratio  (and  thus  change  the  «'"s  in  the  inverse  ratio)  neces- 
sary to  make  these  sums  equal  and  thus  satisfy  condition 
(9).  The  alteration  of  the  z'"s,  all  in  the  same  ratio,  will 


ARCH-RIBS. 


4G7 


not  interfere  with  conditions  (7)  and  (8)  which  are  already 
satisfied. 


381.  Detail  of  Construction  of  Last  Problem.  Symmetrical  Arch- 
Rib  of  Fixed  Ends. — As  an  example  take  a  span  of  the  St. 
Louis  Bridge  (assuming  /constant)  with  "  live  load"  cov- 
ering the  half  span  on  the  left,  Fig.  430,  where  the  vertical 


FIG.  430. 


scale  is  much  exaggerated  for  the  sake  of  distinctness*. 
Divide  into  eight  equal  Js's.  (In  an  actual  example  sixteen 
or  twenty  should  be  taken.)  Draw  a  vertical  through  the 

*  Each  arch-rib  of  the  St.  Louis  bridge  is  a  built  up  or  trussed  rib  of  steel  about  52o 
ft.  span  and  52  ft  rise,  in  the  form  of  a  segment  of  a  circle  .  Its  moment  of  inertia, 
however,'tia  not  strictly  constant,  the  portions  near  each  pier,  of  a  length  equal  to  one 
twelfth  of  the  span,  having  a  value  of  /one-half  greater  than  that  of  the  remainder  of 
the  arc. 


468  MECHANICS   OF    ENGINEERING. 

middle  of  each  Js,  P1}  etc.,  are  the  loads  coming  upon  the 
respective  Js's. 

First,  to  locate  vm,  by  eq.  (6) ;  from  symmetry  it  must 
be  horizontal.  Draw  a  trial  vm  (not  shown  in  the  figure) 
and  if  the  (+z')'s  exceed  the  ( — z')'s  by  an  amount  z0',  the 

true  vm  will  lie  a  height  -  «„'  above  the  trial  vm  (or  be- 
low, if  vice  versa) ;  n  =  the  number  of  Js's. 

Now  lay  off  the  load-line  on  the  right,  (to  scale), 
take  any  convenient  trial  pole  0'"  and  draw  a  correspond- 
ing trial  equil.  polygon  F'"Gm.  In  F'"G'"t  by  §  377, 
locate  a  straight  line  v'"m"  so  as  to  make  2*(z'")=0  and 
™(xz'")=Q  (see  Note  (b)  of  §  377). 

[We  might  now  redraw  F"'G"'  in  such  a  way  as  to  bring 
v'"m'"  into  a  horizontal  position,  thus  :  first  determine  a 
point  n"'  on  the  load-line  by  drawing  0"'ri"  \\  to  v'"m'"  take 
a  new  pole  on  a  horizontal  through  ri",  with  the  same 
H'1',  and  draw  a  corresponding  equil.  polygon  ;  in  the  lat- 
ter v"'m'"  would  be  horizontal.  We  might  also  shift  this 
new  trial  polygon  upward  so  as  to  make  v"'m'''  and  vm 
coincide.  It  would  satisfy  conditions  (7)  and  (8),  having 
the  same  a""s  as  the  first  trial  polygon  ;  but  to  satisfy  con- 
dition (9)  it  must  have  its  z""s  altered  in  a  certain  ratio, 
which  we  must  now  find.  But  we  can  deal  with  the  indi- 
vidual z"''s  just  as  well  in  their  present  positions  in  Fig. 
430.]  The  points  E  and  L  in  vm,  vertically  over  E'"  and 
/,"'  in  v'"m'",  are  now  fixed ;  they  are  the  intersections  of  the 
special  polygon  required,  with  vm. 

The  ordinates  between  v'"m'"  and  the  trial  equilibrium 
polygon  have  been  called  %'"  instead  of  z"\  they  are  pro- 
portional to  the  respective  z"'s  of  the  required  special 
polygon. 

The  next  step  is  to  find  in  what  ratio  the  (z"')'s  need  to 
be  altered  (or  H'"  altered  in  inverse  ratio)  in  order  to  be- 
come the  (s")'s  J  i-e'»  *n  or(ier  to  fulfil  condition  (9),  viz.: 


AKCH-R1BS. 


4G9 


i 

9 

> 

-- 

-^ 

s 

4 

-**. 

x 

N 

r^-  — 
^----" 

^ 

^ 

(«) 

1  —  — 
i 

\^ 

^^ 

^ 

^> 



11.- 
H| 

y—  - 

t«J 

~>j- 

« 

S((yz")=Z\(yz')     .    (9) 

This  may  be  done  pre- 
cisely as  for  the  rib  with 
two  hinges,  but  the  nega- 
tive (O's  must  be  prop- 
erly considered  (§  375) 
See  Fig.  431  for  the  de- 
tail. Negative  z"s  ors""s 
point  upward. 

From  Fig.  431a 


.-.  from  symmetry 


From  Fig.  4316  we  have 


Fio.  431. 

and  from  Fig.  431c 


[The  same  pole  distance  H0  is  taken  in  all  these  construc- 

tions]  .'.   IV)  =  #,(£>  +  &r). 

If,  then,  H0  (fc,+Jfer)  =  ZHJc  condition  (9)  is  satisfied  by  the 
z""8.  If  not,  the  true  pole  distance  for  the  special  equil. 
polygon  of  Fig.  430  will  be 


=. 

2k 

With  this  pole  distance  and  a  pole  in  the  horizontal  through 
»'"  (Fig.  430)  the  force  diagram  may  be  completed  for  the 
required  special  polygon  ;  and  this  latter  may  be  con- 
structed as  follows  :  Beginning  at  the  point  E,  in  vm, 
through  it  draw  a  segment  I!  to  the  proper  ray  of  the  force 
diagram.  In  our  present  figure  (430)  this  "  proper  ray  " 
would  be  the  ray  joining  the  pole  with  the  point  of  meet- 
ing of  P2  and  P3  on  the  load-line.  Having  this  one  seg- 


470  MECHANICS   OF   ENGINEERING. 

ment  of  the  special  polygon  the  others  are  added  in  an 
obvious  manner,  and  thus  the  whole  polygon  completed. 
It  should  pass  through  L,  but  not  0  and  B. 

For  another  loading  a  different  special  equil.  polygon 
would  result,  and  in  each  case  we  may  obtain  the  thrust, 
shear,  and  moment  of  stress  couple  for  any  cross-section  of 
the  rib,  by  §  367.  To  the  stresses  computed  from  these, 
should  be  added  (algebraically)  those  occasioned  by  change 
of  temperature  and  by  shortening  of  the  rib  as  occasioned 
by  the  thrusts  along  the  rib.  These  "  temperature 
stresses,"  and  stresses  due  to  rib-shortening,  will  be  con- 
sidered in  a  subsequent  paragraph.  They  have  no  exist- 
ence for  an  arch-rib  of  three  hinges. 

[NOTE. — If  the  moment  of  inertia  of  the  rib  is  variable 

/  z" 

we  put  —  for  z'  and  —  for  z"  in  equation  (6),  (7),  (8),  and 

(9),  n  having  the  meaning  given  in  the  Note  in  §  379  a, 
which  see ;  and  proceed  accordingly]. 


381a.  Exaggeration  of  Vertical  Dimensions  of  Both  Space  and 
Force  Diagrams. — In  case,  as  often  happens,  the  axis  of  the 
given  rib  is  quite  a  flat  curve,  it  is  more  accurate  (for  find- 
ing M)  to  proceed  as  follows : 

After  drawing  the  curve  in  its  true  proportions  and  pass- 
ng  a  vertical  through  the  middle  of  each  of  the  equal 
z/s's,  compute  the  ordinate  (y)  of  each  of  these  middle  points 
from  the  equation  of  the  curve,  and  multiply  each  y  by 
four  (say).  These  quadruple  ordinates  are  then  laid  off 
from  the  span  upward,  each  in  its  proper  vertical.  Also 
multiply  each  load,  of  the  given  loading,  by  four,  and  then 
with  these  quadruple  loads  and  quadruple  ordinates,  and 
the  upper  extremities  of  the  latter  as  points  in  an  exagge- 
rated rib-curve,  proceed  to  construct  a  special  equilibrium 
polygon,  and  the  corresponding  force  diagram  by  the 
proper  method  (  for  Class  B,  C,  or  D,  as  the  case  may  be) 
for  this  exaggerated  rib -curve. 

The  moment,  Hz,  thus  found  for  any  section  of  the  ex- 


ARCH-RIBS.  471 

aggerated  rib-curve,  is  to  be  divided  by  four  to  obtain  the 
moment  in  the  real  rib,  in  the  same  vertical  line.  To  find 
the  thrust  and  shear,  however,  for  sections  of  the  real  rib, 
besides  employing  tangents  and  normals  of  the  real  rib  we 
must  draw,  and  use,  another  force  diagram,  obtained  from 
the  one  already  drawn  (for  the  exaggerated  rib)  by  re- 
ducing its  vertical  dimensions  (only),  in  the  ratio  of  four 
to  one.  [Of  course,  any  other  convenient  number  besides 
four,  may  be  adopted  throughout.] 

382.  Stress  Diagrams.— Take  an  arch-rib  of  Class  Dt  §  378, 
i.e.,  of  fixed  ends,  and  suppose  that  for  a  given  loading  (in- 
cluding its  own  weight)  the  special 
equil.  polygon  and  its  force  diagram 
have  been  drawn  [§  381].  It  is  re- 
quired  to  indicate  graphically  the 
variation  of  the  three  stress-elements 
for  any  section  of  the  rib,  viz.,  the 
thrust,  shear,  and  mom.  of  stress- 
couple.  I  is  constant.  If  at  any 
point  m  of  the  rib  a  section  is  made,  then  the  stresses  in 
that  section  are  classified  into  three  sets  (Fig.  432).  (See 
§§  295  and  367)  and  from  §  367  eq.  (3)  we  see  that  the  ver- 
tical intercepts  between  the  rib  and  the  special  equil. 
polygon  being  proportional  to  the  products  Hz  or 
moments  of  the  stress-couples  in  the  corresponding  sec- 
tions form  a  moment  diagram,  on  inspection  of  which  we 

can  trace  the  change    in  this  moment,     Hz  =  £?-  ,      and 

hence  the  variation  of  the  stress  per  square  inch,  p^,  (as 
due  to  stress  couple  alone)  in  the  outermost  fibre  of  any 
section  (tension  or  compression)  at  distance  e  from  the 
gravity  axis  of  the  section),  from  section  to  section  along 
the  rib. 

By  drawing  through  0  lines  On'  and  Of  parallel  re- 
spectively to  the  tangent  and  normal  at  any  point  m  of  the 
rib  axis  [see  Fig.  433]  and  projecting  upon  them,  in  turn, 
the  proper  ray  (R,  in  Fig.  433)  (see  eqs.  1  and  2  of  §  367) 


472 


MECHANICS    OF    ENGINEERING. 


we  obtain  the  values  of  the  thrust  and  shear  for  the  sec- 
tion at  m.  When  found  in  this  way  for  a  number  of  points 
along  the  rib  their  values  may  be  laid  off  as  vertical  lines 
from  a  horizontal  axis,  in  the  verticals  containing  the  re- 
spective points,  and  thus  a  thrust  diagram  and  a  shear  dia- 
gram may  be  formed,  as  constructed  in  Fig.  433.  Notice 
that  where  the  moment  is  a  maximum  or  minimum  the 
shear  changes  sign  (compare  §  240),  either  gradually  or 


suddenly,  according  as  the  max.  or  min.  occurs  between 
two  loads  or  in  passing  a  load ;  see  m',  e.  g. 

Also  it  is  evident,  from  the  geometrical  relations  involv- 
e  1,  that  at  those  points  of  the  rib  where  the  tangent-line 
is  parallel  to  the  "  proper  ray  "  of  the  force  diagram,  the 
thrust  is  a  maximum  (a  local  maximum)  the  moment  (of 


ARCH-RIBS  473 

stress  couple)  is  either  a  maximum  or  a  minimum  and  the 
shear  is  zero. 

.From  the  moment,  Hz  =  &?,  &  —  £~ 

may  be  computed.    From  the  thrust  =  Fplt  ^1=th^st>  (F 

=  area  of  cross-section)  may  be  computed.  Hence  the 
greatest  compression  per  sq.  inch  (^i+/>2)  may  be  found  in 
each  section.  A  separate  stress-diagram  might  be  con- 
structed for  this  quantity  (PI+PI).  Its  max.  value  (after 
adding  the  stress  due  to  change  of  temperature,  or  to  rib- 
shortening,  for  ribs  of  less  than  three  hinges),  wherever  it 
occurs  in  the  rib,  must  be  made  safe  by  proper  designing 
of  the  rib.  The  maximum  shear  Jm  can  be  used  as  in  §256 
to  determine  thickness  of  web,  if  the  section  is  I-shaped, 
or  box-shaped.  See  §  295. 

383.  Temperature  Stresses. — In  an  ordinary  bridge  truss 
and  straight  horizontal  girders,  free  to  expand  or  contract 
longitudinally,  and  in  Classes  A  and  B  of  §  378  of  arch- 
ribs,  there  are  no  stresses  induced  by  change  of  tempera- 
ture ;  for  the  form  of  the  beam  or  truss  is  under  no 
constraint  from  the  manner  of  support ;  but  with  the  arch- 
rib  of  two  hinges  (hinged  ends,  Class  C)  and  of  fixed  ends 
(Class  D)  having  immovable  piers  which  constrain  the  dis- 
tance between  the  two  ends  to  remain  the  same  at  all  tem- 
peratures, stresses  called  "  temperature  stresses  "  are  in- 
duced in  the  rib  whenever  the  temperature,  t,  is  not  the 
same  as  that,  ttl,  when  the  rib  was  put  in  place.  These 
may  be  determined,  as  follows,  as  if  they  were  the  only 
ones,  and  then  combined,  algebraically,  with  those  due  to 
the  loading. 

384.  Temperature  Stresses  in  the  Arch-Rib  of  Hinged  Ends.— 
(Class  C,  §  378.)  Fig.  434.  Let  E  and  /be  constant,  with 


474  MECHANICS  OF   ENGINEERING. 

,rt  o,  H       n,  other  postulates  as  in  §  379. 

S^~"~~'\^\  ^e*   ^°    =    temperature    of 

/f       y_z  \      \\  erection,  and  I  =  any  other 

O//H  H \B  temperature;  also   let  I  = 

go  "S  length  of  span  =    OB  (in- 

Fio"~434~~*  variable)  and  r^— co-efficient 

of  linear  expansion  of  the 

material  of  the  curved  beam  or  rib  (see  §  199).  At  tempera- 
ture t  there  must  be  a  horizontal  reaction  H  at  each  hinge 
to  prevent  expansion  into  the  form  O'B  (dotted  curve), 
which  is  the  form  natural  to  the  rib  for  temperature  t  and 
without  constraint.  We  may  /.  consider  the  actual  form 
OB  as  having  resulted  from  the  unstrained  form  O'B  by 
displacing  0'  to  0,  ie.,  producing  a  horizontal  displace- 
ment O'O  =1  (t-tjft. 

But  O'O  =  Ax  (see  §§  373  and  374) ;  (KB.  £'a  tangent 
has  moved,  but  this  does  not  affect  Ax,  if  the  axis  X  is 
horizontal,  as  here,  coinciding  with  the  span  ;)  and  the 
ordinate  y  of  any  point  ra  of  the  rib  is  identical  with  its 
8  or  intercept  between  it  and  the  spec,  equil.  polygon, 
which  here  consists  of  one  segment  only,  viz.  :  OB.  Its 
force  diagram  consists  of  a  single  ray  0\  n' ;  see  Fig.  434. 
Now  (§  373) 

Ax  =  —  CMyds  ;  and  M=Hz  =  in  this  case,  Hy 

, ,.     .  x        H   /»B2  T      (  hence  for  graphics,  and 
.-.  I  (t-Q  y=—  ffds  ;  \  ,    X    *     ,  ' 


e 

.     .      (1) 

From  eq.  (1)  we  determine  //,  having  divided  the  rib-curve 
into  from  twelve  to  twenty  equal  parts  each  called  Js . 

For  instance,  for  wrought  iron,  t  and  £0,  being  expressed 
in  Fahrenheit  degrees,  y  =  0.0000066.  If  E  is  expressed 
in  Ibs.  per  square  inch,  all  linear  quantities  should  be  in 
inches  and  H  will  be  obtained  in  pounds. 

2ly*  may  be  obtained  by  §  375,  or  may  be  computed.  H 
being  known,  we  find  the  moment  of  stress-couple  =  Hyt 


ARCH-RIBS.  475 

at  any  section,  while  the  thrust  and  shear  at  that  section 
are  the  projections  of  H,  i.e.,  of  O^n'  upon  the  tangent  and 
normal.  The  stresses  due  to  these  may  then  be  determined 
in  any  section,  as  already  so  frequently  explained,  and 
then  combined  with  those  due  to  loading. 

385.  Temperature  Stresses  in  the  Arch-Ribs  with  Fixed  Ends.— 
See  Fig.  435.  (Same  postulates  as  to  symmetry,  E  and  1 
constant,  etc.,  as  in  §  380.)  t  and  t0  have  the  same  meaning 
as  in  §  384. 

Here,  as  before,  we 
consider  the  rib  to 
have  reached  its  ac- 
tual form  under  tem- 
perature t  by  having 
had  its  span  forcibly  , 
shortened  from  the  _AJ 

length     natural     to       °  "^ * — FiiTtts 

temp,  t,  viz.:    O'B', 

to  the  actual  length  OB,  which  the  immovable  piers  compel 
it  to  assume.  But  here,  since  the  tangents  at  0  and  B  are 
to  be  the  same  in  direction  under  constraint  as  before,  the  two 
forces  H,  representing  the  action  of  the  piers  on  the  rib, 
must  be  considered  as  acting  on  imaginary  rigid  prolonga- 
tions at  an  unknown'distance  d  above  the  span.  To  find 
H  and  d  we  need  two  equations. 

From  §  373  we  have,  since  M=Hz=H(y-d), 


.  (2) 
or,  graphically,  with  equal  ^/s's 

'  (3) 


Also,  since  there  has  been  no  change  in  the  angle  between 
end-tangents,  we  must  have,  from  §  374, 


476  XECHAKICS  or  ENGINEERING. 

or   for  graphics,  with  equal  Av's,  -ly  =  nd     .     .          .     (4) 

in  which  n  denotes  the  number  of  Js's.  From  (4)  we 
determine  d,  and  then  from  (3)  can  compute  H.  Drawing 
the  horizontal  F  G,  it  is  the  special  equilibrium  polygon 
(of  but  one  segment)  and  the  moment  of  the  stress-couple 
at  any  section  =  Hz,  while  the  thrust  and  shear  are  the 
projections  of  H=  0\n'  on  the  tangent  and  normal  respect- 
ively of  any  point  m  of  rib. 

For  example,  in  one  span,  of  550  feet,  of  the  St.  Louis 
Bridge,  having  a  rise  of  55  feet  and  fixed  at  the  ends,  the 
force  H  of  Fig.  435  is  =  108  tons,  when  the  temperature  is 
80°  Fahr.  higher  than  the  temp,  of  erection,  and  the  en- 
forced span  is  3j^  inches  shorter  than  the  span  natural  to 
that  higher  temperature.  Evidently,  :.f  the  actual  temp- 
erature t  is  lower  than  that  tia  of  erection,  Hmnst  act  in  a 
direction  opposite  to  that  of  Figs.  435  and  434,  and  the 
*'  thrust  "  in  any  section  will  be  negative,  i.e.,  a  pull. 

386.  Stresses  Due  to  Rib-Shortening  —In  §  369,  Fig.  407,  the 
shortening  of  the  element  AE  to  a  length  A'E,  due  to  the 
uniformly  distributed  thrust,  p\F,  was  neglected  as  pro- 
ducing indirectly  a  change  of  curvature  and  form  in  the 
rib  axis  ;  but  such  will  be  the  case  if  the  rib  has  less  than 
three  hinges.  This  change  in  the  length  of  the  different 
portions  of  the  rib.  curve,  may  be  treated  as  if  it  were  due 
to  a  change  of  temperature.  For  example,  from  §  199  we 
«ee  that  a  thrust  of  50  tons  coming  upon  a  sectional  area 
of  F  =  10  sq.  inches  in  an  iron  rib,  whose  material  has  a 
modulus  of  elasticity  =  E  =  30,000,000  Ibs.  per  sq.  inch, 
and  a  coefficient  of  expansion  #  =  .0000066  per  degree 
Fahrenheit,  produces  a  shortening  equal  to  that  due  to  a 
fall  of  temperature  (t0—t)  derived  as  follows  :  (See  §  199) 
(units,  inch  and  pound) 

ft  -t}=     P  -  100'000  -50° 

10x30,000,000x.0000066~' 


Fahrenheit. 

Practically,   then,   since   most    metal   arch  bridges   of 
cla-5S3s  C  and  D  are  rather  flat  in  curvature,  and  the  thrusts 


AKCH-KIBS.  477 

due  to  ordinary  modes  of  loading  do  not  vary  more  than  20 
or  30  per  cent,  from  each  other  along  the  rib,  an  imagin- 
ary fall  of  temperature  corresponding  to  an  average  thrust 
in  any  case  of  loading  may  be  made  the  basis  of  a  con- 
struction similar  to  that  in  §  384  or  §  385  (according  as  the 
ends  are  hinged,  or  fixed)  from  which  new  thrusts,  shears, 
and  stress-couple  moments,  may  be  derived  to  be  combin- 
ed with  those  previously  obtained  for  loading  and  for 
change  of  temperature. 

387  Resume  —  It  is  now  seen  how  the  stresses  per  square 
inch,  both  shearing  and  compression  (or  tension)  may  be 
obtained  in  all  parts  of  any  section  of  a  solid  arch-rib  or 
curved  beam  of  the  kinds  described,  by  combining  the  re- 
sults due  to  the  three  separate  causes,  viz.:  the  load, 
change  of  temperature,  and  rib-shortening  caused  by  the 
thrusts  due  to  the  load  (the  latter  agencies,  however,  com- 
ing into  consideration  only  in  classes  C  and  D,  see  §  378). 
That  is,  in  any  cross-section,  the  stress  in  the  outer  fibre 
is,  [letting  T},(,  T^f,  Th'",  denote  the  thrusts  due  to  the 
three  causes,  respectively,  above  mentioned  ;  (Hz)',  (Hz)", 
(Hz)'",  the  moments] 

±  (B*)"±  (Ifr)'"       .     .     .     (1) 


Le.,  Ibs.  per  sq.  inch  compression  (if  those  units  are  used). 
The  double  signs  provide  for  the  cases 
where  the  stresses  in  the  outer  fibre,  due 
c  to  a  single  agency,  may  be  tensile.  Fig. 
436  shows  the  meaning  of  e  (the  same 
used  heretofore)  /is  the  moment  of  in- 
ertia of  the  section  about  the  gravity 
axis  (horizontal)  C.  F  =  area  of  cross- 
section.  [0!  =  e  ;  cross  section  symmet- 
rical about  C].  For  a  given  loading  we 
may  find  the  maximum  stress  in  a  given  rib,  or  design  the 
rib  so  that  this  maximum  stress  shall  be  safe  for  the  ma- 
terial employed.  Similarly,  the  resultant  shear  (total,  not 


478 


MECHANICS   OF    ENGINEERING. 


per  sq.  inch)  =  J'  ±  J"  ±  J'"  is  obtained  for  any  section 
to  compute  a  proper  thickness  of  web,  spacing  of  rivets, 
etc. 

388  The  Arch-Truss,  or  braced  arch..  An  open-work 
truss,  if  of  homogeneous  design  from  end  to  end,  may  be 
treated  as  a  beam  of  constant  section  and  constant  moment 
of  inertia",  and  if  curved,  like  the  St.  Louis  Bridge  and  the 
Coblenz  Bridge  (see  §  378,  Class  D),  may  be  treated  as  an 
arch-rib.*  The  moment  of  inertia  may  be  taken  as 


where  FY  is  the  sectional  area  of  one  of  the  pieces  (I  to  the 
curved  axis  midway  between  them,  Fig.  437,  and  h  =  dis- 
tance between  them. 


FIG.  438.  FIO.  437. 

Treating  this  curved  axis  as  an  arch-rib,  in  the  usual 
way  (see  preceding  articles),  we  obtain  the  spec,  equil.  pol. 
and  its  force  diagram  for  given  loading.  Any  plane  ~]  to 
the  rib-axis,  where  it  crosses  the  middle  m  of  a  "  web- 
member,"  cuts  three  pieces,  A,  B  and  C,  the  total  com- 

*The  St  Louis  Bridge  u  not  strictly  of  constant  moment  of  Inertia,  being  somewhat 
strengthened  near  each  pier 


ARCH-RIBS. 


479 


pressions  (or  tensions)  in  which  are  thus  found  :  For  the 
point  ra,  of  rib-axis,  there  is  a  certain  moment  =  Hz,  a 
thrust  —  rh,  and  a  shear  =  J,  obtained  as  previously  ex- 
plained. We  may  then  write  Psin/3  =  J (1) 

and  thus  determine  whether  P  is  a  tension  or  compres- 
sion ;  then  putting  P'+P"  ±  P  cos  ft  =  Th 2 

(in  which  P  is  taken  with  a  plus  sign  if  a  compression,  and 
minus  if  tension);  and 


(3) 


we  compute  P1  and  P",  which  are  assumed  to  be  both  com- 
pressions here.  /?  is  the  angle  between  the  web  member 
and  the  tangent  to  rib-axis  at  m,  the  middle  of  the  piece. 
See  Fig.  406,  as  an  explanation  of  the  method  just 
adopted. 


HORIZONTAL  STRAIGHT  GIRDERS. 

389,  Ends  Tree  to  Turn. — This  corresponds  to  an  arch- 
rib  with  hinged  ends,  but  it  must  be  understood  that  there 
is  no  hindrance  to  horizontal  motion.  (Fig.  439.)  In 


Fio   439. 


treating  a  straight  beam,  slightly  bent  under  vertical  forces 
only  (as  in  this  case  with  no  horizontal  constraint),  as  a 


480  MECHANICS   OF   ENGINEERING. 

particular  case  of  an  arch-rib,  it  is  evident  that  since  the 
pole  distance  must  be  zero,  the  special  equil.  polygon  will 
have  all  its  segments  vertical,  and  the  corresponding  force 
diagram  reduces  to  a  single  vertical  line  (the  load  line). 
The  first  and  last  segments  must  pass  through  A  and  B 
(points  of  no  moment)  respectively,  but  being  vertical  will 
not  intersect  Pt  and  P2 ;  i.e.,  the  remainder  of  the  special 
equilibrium  polygon  lies  at  an  infinite  distance  above  the 
span  AE.  Hence  the  actual  spec,  equil.  pol.  is  useless. 

However,  knowing  that  the  shear,  J,  and  the  moment 
M  (of  stress  couple)  are  the  only  quantities  pertaining  to 
any  section  ra  (Fig.  439)  which  we  wish  to  determine  (since 
there  is  no  thrust  along  the  beam),  and  knowing  that  an 
imaginary  force  H'y  applied  horizontally  at  each  end  of  the 
beam,  would  have  no  influence  in  determining  the  shear 
and  moment  at  m  as  due  to  the  new  system  of  forces,  we 
may  therefore  obtain  the  shears  and  moments  graphically 
from  this  new  system  (viz.:  the  loads  Plt  etc.,  the  vertical 
reactions  V  and  VM  and  the  two  equal  and  opposite  /T's). 
[Evidently,  since  H1'  has  no  moment  about  the  neutral 
axis  (or  gravity  axis  here),  of  m,  the  moment  at  m  will  be 
unaffected  by  it ;  and  since  H"  has  no  component ""]  to  the 
beam  at  m,  the  shear  at  m  is  the  same  in  the  new  system 
of  forces,  as  in  the  old,  before  the  introduction  of  the 
tf"s.] 

Hence,  lay  off  the  load-line  1  .  .  2 .  .  3,  Fig.  439,  and  con- 
struct an  equil.  polyg.  which  shall  pass  through  A  and  B 
and  have  any  convenient  arbitrary  H"  (force)  as  a  pole 
distance.  This  is  dona  by  first  determining  n  on  the  load- 
line,  using  the  auxiliary  polygon  A'a'F,  to  a  pole  (7  (arbi- 
trary), and  drawing  O'n1  \\  to  A'B'.  Taking  O1'  on  a  hori- 
zontal through  n,  making  0''n'=H",  we  complete  the 
force  diagram,  and  equil.  pol.  AaB.  Then,  z  being  the  ver- 
tical intercept  between  m  and  the  equil.  polygon,  we  have: 
Moment  at  m~M=H"z  (or=ITz'  also),  and  shear  at  m,  or 
•/,— 2  .  .  n't  i.e.,  —  projection  of  the  proper  ray  J?2>  or 
0"  .  .  2,  upon  the  vertical  through  m.  Similarly  we  ob- 
tain M  and  J  at  any  other  section  for  the  given  load.  (See 


ARCH-KI13S ;   SPECIAL   CASE  ;    STRAIGHT. 


481 


§§  329,  337  and  367).     The  moment  of  inertia  need  not  be 
constant  in  this  case. 

390.  Straight  Horizontal  Prismatic  Girder  of  Fixed  Ends  at  Same 
Level. — No  horizontal  constraint,  hence  no  thrust.  /  con- 
stant. Ends  at  same  level,  with  end-tangents  horizontal. 
We  may  consider  the  whole  beam  free  (cutting  close  to  the 
walls)  putting  in  the  unknown  upward  shears  J0  and  Jw 


i       2 


and  the  two  stress  couples  of  unknown  moments  JH0  and 
M,,  at  these  end  sections.  Also,  as  in  §  388,  an  arbitrary 
H"  horizontal  and  in  line  of  beam  at  each  extremity.  Now 
(See  Fig.  33)  the  couple  at  0  and  the  force  H"  are  equiv- 
alent to  a  single  horizontal  H"  at  an  unknown  vertical  dis- 
tance c  below  0 ;  similarly  at  the  right  hand  end.  The 
special  polygon  FG  is  to  be  determined  for  this  new  sys- 
tem, since  the  moment  and  shear  will  be  the  same  at  any 
section  under  this  new  system  as  under  the  real  system. 
The  conditions  for  determining  it  are  as  follows  :  Since 
the  end-tangents  are  fixed,  -Mds=Q  .*.  JTJ«//s=0  and  since 


482  MECHANICS   OF   ENGINEERING. 

O's  displacement  relatively  to  J5's  tangent  is  zero  we  nave 
2Mxds=0  .-.  ZH"zxAs=0  .'.  2'xzJs=0.  See  §  374.  Hence 
for  Equal  4s\  l'(z)=Q  and  I(xz)=0.  Now  for  any  pole  0'" 
draw  an  equil.  pol.  F'"G'"  and  in  it  (by  §  377;  see  Note) 
locate  v'"m,'"  so  as  to  make  2'(3"')=0  and  2"(2*'")=0. 
Draw  verticals  through  the  intersections  E'"  and  L"'t  to 
determine  E  and  Z  on  the  beam,  these  are  the  points  of 
inflection  (i.e.,  of  zero  moment),  and  are  points  in  the  re- 
quired special  polygon  FG. 

Draw  0"'n"  II  to  v'"m'"  to  fix  n".  Take  a  pole  0"  on 
the  horizontal  through  n",  making  0/'y,"=H"  (arbitrary), 
draw  the  force  diagram  0"  1234  and  a  corresponding 
equilibrium  polygon  beginning  at  E.  It  should  cut  L, 
and  will  fulfil  the  two  requirrnents  2*(z)=Q  and  ~*(xz)=Q, 
with  reference  to  the  axis  of  the  beam  O'B'.  The  moment  of 
the  stress-couple  at  any  section  m  will  be  M=H"z,  and  the 
shear  J  =  the  projection  of  the  "  proper  ray  "  of  the  force 
diagram  0"  .  .  1,  2,  etc.,  upon  the  vertical  (not  in  the  trial 
diagram  0'".  .  1,  2,  etc.).  As  far  as  the  moment  is  concern- 
ed the  trial  polygon  F'"  G'"  will  serve  as  well  as  the  special 
polygon  FG  ;  i.Q.,M=H'"z'"  as  well  as  H"zt  H'"  being  the 
pole-distance  of  0'"  ;  but  for  the  shear  we  must  use  the 
rays  of  the  final  and  not  the  trial  diagram. 

The  peculiarity  of  this  treatment  of  straight  beams, 
considered  as  a  particular  case  of  curved  beams,  consists 
in  the  substitution  of  an  imaginary  system  of  forces  in- 
volving the  two  equal  and  opposite,  and  arbitrary  H's,  for 
the  real  system  in  which  there  is  no  horizontal  force  and 
consequently  no  "  special  equilibrium  polygon,"  and  thus 
determining  all  that  is  desired,  i.e.,  the  moment  and  shear 
at  any  section. 

In  the  polygon  FG  the  student  will  recognize  the  "  mo- 
ment-diagram "  of  the  problems  in  Chaps.  Ill  and  IV. 

He  will  also  see  why  the  shear  is  proportional  to  the 

slope  -—  of  the  moment  curve   in   those  chapters.     For 

ax 

example,  the  "  slope  "  of  the  second  segment  of  the  poly- 
gon FG,  that  segment  being  II  to  0"  2,  is 


ARCH-KIBS  ;    SPECIAL   CASE  ;    STRAIGHT.  483 

tang,  of  angle  20'V'=2n"-f-0'V=shear  -r-  H" 

and  similarly  for  any  other  segment ;  i.e.,  the  tangent  of 
the  inclination  of  the  "  moment  curve,"  or  line,  is  propor- 
tional to  the  shear. 

It  is  also  interesting  to  notice  with  the  present  problem 
of  a  straight  beam,  that  in  the  conditions 

Jf(sJs)=0  and  2(zAs)x=Q, 

for  locating  the  polygon  FG,  each  4s  is  1  to  its  z,  and 
that  consequently  each  zAs  is  the  area  of  a  small  vertical 
strip  of  area  between  the  beam  and  the  polygon,  and 
(zAs)x  is  the  "  moment"  of  this  strip  of  area,  about  0'  the 
origin  of  x.  Hence  these  conditions  imply  ;  first,  that  the 
area  EWL  between  the  polygon  and  the  axis  of  the  beam 
on  one  side  is  equal  to  that  (CfFE+LB'G)  on  the  other 
side,  and,  secondly,  that  the  centre  of  gravity  of  EWL  lies 
in  the  same  vertical  as  that  of  O'FE  and  LB'  G  combined. 
Another  way  of  stating  the  same  thing  is  that,  if  we  join 
FG,  the  area  of  the  trapezoid  FO'B1  G  is  equal  to  that  of  the 
figure  FEWLG,  and  their  centres  of  gravity  lie  in  the  same 
vertical.  A  corresponding  statement  may  be  made  (if  we  join 
F"'G"'^  for  the  trapezoid  F"'v'"m"'G"\  and  figure 


484: 


MECHANICS   OF   EKGLNEEKIXG. 


CHAPTER  XII. 


GRAPHICS  OF  CONTINUOUS  GIRDERS. 


[MAINLY  DUE  TO  PROF.  MOHR.  OF  AIX-LA-CHAPELLE  ] 

391.  The  Elastic  Curve  of  a  Horizontal  Loaded  Beam,  Homoge« 
neous  and  Originally  Straight  and  Prismatic,  is  an  Equilibrium 
Polygon,  whose  "load-line"  i<s  vertical 
and  consists  of  the  successive  products 
Mdx  [treated  as  if  they  were  loads  A 
each  applied  through  the  middle  of 
its  proper  dx~\,  and  whose  "pole  dis- 
tance "  is  EL  Fig  441  (exaggerated). 

Let  AO  and  OC  be  any  two  con- 
secutive  equal  elements  of  a  very 
flat  elastic  curve  (as  above  described). 
Prolong  AO  to  cut  NC.  Then  from 
§  231,  eq.  (7),  we  have 


to  --  ET       •••• 

where  d2y=DC;and,  hence,  if  a  triangle  (Fig.  442)  O'D'C', 
be  formed  with  O'D'  II  to  OD,  O'C'  II  to  OC,  and  D'C'  ver- 
tical, while  its  (horizontal)  altitude  O'n  is  made  equal,  by 
scale,  to  El  of  the  beam,  then  from  the  similarity  of  the 
triangle  ODC  and  O'D'C'  and  the  proportion  in  eq.  (1)  we 
see  that  D'C'  must  represent  the  product  Mdx  on  the  same 
scale  by  which  O'n  represents  El,  31  is  the  moment  of 


CONTINUOUS  GIRDER  BY  GRAPHICS.  485 

the  stress-couple  at  the  section  whose  neutral  axis  is  pro- 
jected in  0. 

Similarly,  if  Mf  is  the  moment  of  the  stress-couple  at 
C,  and  we  draw  O'F',  II  to  CF,  C'F'  must  represent  M'dx 
(on  same  scale).  It  is  therefore  apparent  that  the  line 
AOCF  bears  to  the  figure  O'D'C'F'  the  same  relation 
which  an  equilibrium  polygon  (for  vertical  forces)  does  to 
its  force  diagram,  the  "  loads  "  of  the  force-diagram  being 
the  successive  values  of  Mdx  laid  off  to  scale,  while  its 
"  pole-distance  "  is  El  laid  off  on  the  same  scale.  [As  if 
Mdx  and  M'dx  were  loads  suspended  at  0  and  C  respec- 
tively.] 

Practically,  since  any  actual  elastic  curve  is  very  flat, 
and  since  a  change  of  pole-distance  will  change  all  verti- 
cal dimensions  of  the  equilibrium  polygon  in  an  inverse 
equal  ratio,  we  may  exaggerate  the  vertical  dimensions  of 
the  elastic  curve  by  choosing  a  pole  distance  smaller  than 
El  in  any  convenient  ratio,  n.  Any  deflection  in  the  elas- 
tic curve  thus  obtained  will  be  greater  than  its  true  value 
in  the  same  ratio  n. 

Graphically,  in  order  to  draw  exaggerated  elastic  curves 
according  to  this  principle,  we  obtain  approximate  results 
by  dividing  the  length  of  the  beam  into  a  number  of  equal 
Ac's,  draw  verticals  through  the  middles  of  the  Jx's  as 
"  force-verticals,"and  lay  off  as  a  "  load-line  "  to  any  con- 
venient scale  the  corresponding  values  of  MAx  in  their 
proper  order. 

The  quality  of  the  product  MJx  is  evidently  (length)2  x 
force,  and  with  the  foot  and  pound  as  units  such  a  product 
may  be  called  so  many  (sq.  ft.)  (Ibs.).  It  will  be  noticed 
that  these  products  (MJx)  are  proportional  to,  and  maybe 
represented  by,  the  areas  of  the  corresponding  vertical 
strips  of  the  "  moment-diagram "  proper  to  the  case  in 
hand,  These  strips  together  make  up  the  " moment-area" 
as  it  may  be  called,  lying  between  the  moment  curve  and 
its  horizontal  axis  (which  is  the  axis  of  the  beam  itself, 
according  to  §§  389  and  390). 


480  MECHANICS    OF    ENGINEERING. 

392.  Mohr's  Theorem. — The  principle  of  the  previous 
paragraph  may  therefore  be  enunciated  as  follows  :  That 
just  as  the  moment  curve  (of  a  straight  prismatic  horizontal 
beam)  between  two  consecutive  supports  is  an  equilibrium  poly- 
gon for  the  loading  between  those  supports,  so  also  is  the  elastic 
curve  itself  an  equilibrium  polygon  for  the  "  moment-area  "  con- 
sidered as  a  loading. 

In  dealing  with  the  moment-curve  of  a  single  span  the 
pole  distance  is  arbitrary  (§§  389  and  390),  but  the  position 
of  the  pole  relatively  to  the  load  line  in  other  respects,  and 
the  location  of  the  moment-curve  (equil.-pol.)  relatively  to 
the  beam  (considered  to  be  still  straight  for  this  purpose), 
depend  on  whether  the  beam  simply  rests  on  the  two  sup- 
ports, without  projecting  beyond ;  or  is  built  in,  and  at 
what  angles  ;  or  as  with  a  continuous  girder,  on  the  inclina- 
tion of  the  tangent-lines  at  the  supports,  as  influenced  by 
the  presence  of  loads  on  all  the  spans,  and  on  whether  all 
supports  are  on  the  same  level  or  not. 

For  example,  in  §  389,  for  a  single  span,  the  ends  of 
beam  being  simply  supported  without  overhanging,  the 
pole  0"  must  be  on  a  horizontal  through  nf,  and  the  mo- 
ment curve  must  pass  through  the  extremities  A  and  B  of 
the  beam,  thus  giving  a  "  moment-area  "  lying  entirely  on 
one  side  of  the  beam  (or  axis  from  which  the  moment  or- 
dinates,  z,  are  to  be  measured)  ;  whereas,  in  §  390,  also  a 
single  span,  where  the  ends  of  the  beam  are  built  in  hori- 
zontally and  at  the  same  level,  the  pole  must  be  taken  on 
the  horizontal  through  TO",  and  the  moment-curve  FEWLGr 
must  intersect  the  beam  in  the  points  E  and  L  (E,  L,  and 
n"  being  found  as  prescribed  in  that  problem),  and  thus 
lies  partly  above  and  partly  below  the  beam.  It  will  be 
necessary,  later,  to  distinguish  the  upper  and  lower  parts 
of  the  moment-area  as  positive  and  negative. 

In  drawing  the  equilibrium  polygon  which  constitutes 
the  actual  elastic  curve,  however,  and  hence  making  use  of 
the  successive  small  vertical  strips  of  the  moment-area, 
(when  found)  as  if  they  were  loads,  to  form  a  load-line  ac- 
cording to  a  convenient  scale,  the  pole  distance  is  not  ar- 


CONTINUOUS  GIRDER  BY  GRAPHICS. 


487 


bitrary  but  must  be  =  -£77  on  the  same  scale.  Still,  since 
for  convenience  we  must  always  greatly  exaggerate  the 
vertical  scale  of  the  elastic  curve,  we  may  make  the  pole 
distance  —  JEI-^-n  and  thus  obtain  an  elastic  curve  whose 
vertical  dimensions  are  n  times  as  large  as  those  of  the  real 
curve ;  while  the  position  of  the  pole  will  depend  in  the 
direction  of  the  tangent  lines  at  the  extremities  of  the  span. 
An  example  will  now  be  given. 

393  Example  of  an  Elastic  Curve  (Beam  Prismatic)  Drawn  as 
an  Equilibrium  Polygon  Supporting  the  Moment-Area  as  Loading. 
— Let  the  beam  be  simply  supported  at  its  extremities  (at 
the  same  level),  and  bear  a  single  eccentric  load  P,  Fig. 
443,  its  own  weight  being  neglected.  The  moment-area 


consists  of  a  triangle  AGE  [see  first  part  of  §  260,  or  use 
the  graphic  method  of  §  389,  thus  utilizing  a  force  diagram 
012.],  its  altitude  being  the  moment  represented  by  the 

ordinate  CD  and  having  a  value  =  ~p.  Hence  the  total 
moment-area  =  l/2  base  AB  x  mom.  CD 


X       I  = 


488  MECHANICS   OF   ENGINEERING. 

Divide  AB  into  (say)  eight  equal  Ax's,  (eight  are  rather 
few  in  practice ;  sixteen  or  twenty  would  be  better)  and 
draw  a  vertical  through  the  middle  of  each.  Note  the 
portion  of  each  of  these  vertical  intercepts  between  the 
axis  of  the  beam  and  the  moment-curve  AGE.  The  pro- 
ducts H  Ax  for  the  different  subdivisions  are  proportional 
to  these  intercepts,  since  all  the  ^x's  are  equal,  and  are  the 
respective  moment-areas  of  the  4x's. 

Treating  these  products  as  if  they  were  loads,  we  lay  off 
the  corresponding  intercepts  (or  their  halves,  or  quarters, 
or  other  convenient  fractional  part  or  multiple),  from  E 
downwards  to  form  a  vertical  "  load-line,"  beginning  with 
fche  left-hand  intercept  and  continuing  in  proper  order. 

As  to  what  scale  this  implies,  we  determine  by  dividing 
the  total  moment-area  thus  laid  off,  viz.  :  *£  Plh,  say  in 
(sq.  in.)  (Ibs.),  by  the  length  of  EF  in  inches,  thus  obtain- 
ing the  number  of  (sq.  in.)  (Ibs.)  which  each  linear  inch  of 
paper  represents. 

On  this  scale  the  number  of  inches  of  paper  required  to 
represent  the  El  oi  the  beam  is  so  enormous,  that  in  its 
stead  we  use  the  nth  portion,  n  being  an  arbitrary  abstract 
number  of  such  magnitude  as  to  make  El  -4-  n  a  con- 
venient pole-distance,  TS. 

The  proper  position  of  the  pole  0'  on  the  vertical  TWt 
is  fixed  by  the  fact  that  the  elastic  curve,  beginning  at  At 
must  terminate  in  B,  at  the  same  level  as  A.  Hence, 
assuming  any  trial  pole  as  0",  and  drawing  rays  in  the 
usual  manner  (except  that,  as  henceforth,  the  pole  is  taken 
on  the  right  of  the  "  load-line,"  instead  of  on  the  left,  so 
as  to  make  the  resulting  equilibrium  polygon  correspond 
in  direction  of  curvature  to  the  actual  elastic  curve),  we 
draw  the  corresponding  equilibrium  polygon  A"B".  De- 
termining n'  by  drawing  through  0"  a  line  II  to  the  right 
line  A"0",  we  draw  a  horizontal  through  n'  to  intersect 
TJVin  0',  the  required  pole. 

With  0'  as  pole  a  new  equilibrium  polygon  begun  at  A' 
will  terminate  in  B'  and  its  vertical  ordinates  will  be  n 
times  as  great  as  those  of  corresponding  points  on  the 


CONTINUOUS  GIRDER  BY  GRAPHICS.  489 

actual  elastic  curve  AB.  The  same  relation  holds  between 
the  tangents  of  the  angle  of  inclination  to  the  horizontal 
at  corresponding  points  (i.e.,  those  in  same  vertical)of  the 
two  curves. 

394.  Numerical  Case  of  Foregoing  Example. — With  the  inch 
and  pound  as  units,  let  P  =  120  Ibs.,  li  =  40  in.,  L  =  80  in. 
while  the  prismatic  beam  is  of  timber  having  a  modulus 
of  elasticity  E  =  2,000,000  Ibs.  per  sq.  inch,  and  is  rectan- 
gular in  section,  being  2  in.  wide  and  4  in.  high,  so  that 
(its  width  being  placed  horizontally)  the  moment  of  iner- 
tia of  the  section  is  /  =  */„  btf  =  yi2x2x64  =  10^  bi- 
quadratic inches  (§  90.)  Required  the  maximum  deflection. 
Adopting  1:20  as  the  scale  for  distances  (i.e.,  one  linear 
inch  of  paper  to  twenty  inches  of  actual  distance)  we  make 
the  horizontal  AB  6  in.  long,  Fig.  443,  and  AD  2  in.,  tak- 
ing a  point  C  at  convenience  in  the  vertical  through  D, 
and  joining  AC  and  CB,  thus  determining  the  moment- 
diagram  for  this  case.  [As  to  what  pole  distance,  H,  is 
implied  in  this  selection  of  (7,  is  immaterial  in  this  simple 
case  of  a  single  load ;  hence  we  do  not  draw  the  corre- 
sponding force-diagram  at  all.]  We  divide  AB  into  eight 
equal  parts  and  draw  a  vertical  through  the  middle  of 
each.  The  intercepts,  in  these  verticals,  between  AB  and 
the  broken  line  ACB  we  lay  off  from  E  toward  F  as  pre- 
scribed in  §  393.  (By  taking  DC  small  enough  the  line  EF 
will  not  be  inconveniently  long.) 

Suppose  this  length  EF  measures  6.4  inches  on  the 
paper  (as  in  the  actual  draft  by  the  writer).  Since  it  rep- 
resents a  moment-area  of 

^PZ,?2=>£xl20x40x80=192,000  (sq.  in.)  (Ibs.),  the  scale 
of  our  "moment-area-diagram,"  as  we  may  call  it,  must  be 
192,000^6.4=30,000  (sq.  in.)  (Ibs.)  per  linear  inch  of  paper. 

Now  JET=21,333,333  (sq.  in.)  (Ibs.),  which  on  the  above 
scale  would  be  represented  by  711  linear  inches  of  paper. 
With  tt=100,  however,  we  lay  off  ST=EI+n  =  7.1l  inches 
of  paper  as  a  pole  distance,  and  with  a  trial  pole  0"  in 


490  MECHANICS   OF    ENGINEERING. 

the  vertical  TW  draw  the  trial  equilibrium  polygon  or 
elastic  curve  A"B",  and  with  it  determine  n',  then  the  final 
polygon  A'B'  as  already  prescribed.  In  A'B'  we  find  the 
greatest  ordinate,  NK,  to  measure  0.88  inches  of  paper, 
which  represents  an  actual  distance  of  0.88  x  20  =  17.6 
inches  But  the  vertical  dimensions  of  the  exaggerated 
elastic  curve  A'B'  are  rc=100  times  as  great  as  those  of  the 
actual,  hence  the  actual  max.  deflection  is  d=17.6-^w=0.176 
in.  [This  maximum  deflection  could  also  be  obtained  from 
the  oblique  polygon  A"B"  whose  vertical  dimensions  are 
equal  to  those  of  A'B'.  By  the  formula  of  §  235  we  ob- 
tain d=0.174  inches.] 

395.  Direction  of  End-Tangents  of  Elastic  Curve  in  the  Foregoing 
Problem. — As  an  illustration  bearing  on  subsequent  work 
let  us  suppose  that  the  only  result  required  in  §  394  is 
tan  «0,  i.e.,  the  tangent  of  the  angle  B'A'T',  which  the 
tangent-line  B'T'  to  the  elastic  curve  at  the  extremity  A\ 
Fig.  343,  makes  with  the  horizontal  line  A'B',  (tan.  «0  is 
called  the  "slope,"  at  A.}  Let  7?'$'  be  the  tangent-line  at 
B'.  These  two  "  end-tangents  "  are  parallel  respectively  to 
EO'  and  FO',  and  intersect  at  some  point  R.  Now  since 
A' KB'  is  an  equilibrium  polygon  sustaining  an  imaginary 
set  of  loads  represented  by  the  successive  vertical  strips  of 
the  moment-area  ACB,  the  intersection  It  must  lie  in  the 
vertical  containing  the  centre  of  gravity,  U,  of  that  mo- 
ment-area [§  336 j . 

Hence,  if  the  vertical  containing  U  is  known  in  advance, 
or,  as  in  the  present  case,  is  easily  constructed  without 
making  the  strip-subdivision  of  §  394,  we  may  determine 
the  end-tangents  very  briefly  by  considering  the  whole 
moment-area,  M.A.,  (considered  as  a  load)  applied  in  the 
vertical  through  Ut  as  follows  : 

Since  ACB  is  a  triangle,  we  find  U  by  bisecting  AB  in 
X,  joining  CX,  and  making  XU  —  ^  XC,  and  then  draw  a 
vertical  through  U.  Laying  off  EF=6A  inches  [so  as  to 
represent  a  moment-area  of  192,000  (sq.  in.)  (Ibs.)  on  a 
scale  of  30,000  (sq.  in.)  (Ibs.)  per  linear  inch  of  paper], 


CONTINUOUS    GIRDER  BY   GRAPHICS.  491 

and  making  ST=7.ll  inches  as  before,  we  assume  a  trial 
pole  0"  on  TW,  draw  the  two  rays  0"E  and  0"F,  construct 
the  corresponding  trial  polygon  of  two  segments  A"R"B", 
for  the  purpose  of  finding  n'.  With  a  pole  0'  on  TW&nd 
on  a  level  with  n'  we  draw  the  two  rays  O'E  and  O'F,  and 
the  corresponding  segments  A'R,  and  RB'.  (B'  should  be 
on  a  level  with  A',  as  a  check.)  These  two  segments  are 
the  end-tangents  required. 
We  have,  therefore, 


tan  «0  =  'BTTr  -=-  AtW^ 

In  the  present  numerical  problem  we  find  B'T'  to  mea- 
sure 3  in.  of  paper,  i.e.,  60  in.  of  actual  distance  for  the 
exagg.  elastic  curve,  and  therefore  0.60  in.  in  the  real  elas- 
tic curve  (with  n  =  100) 


It  is  now  evident  that  the  position  and  direction  of  the 
end-tangents  of  the  elastic  curve  lying  beticeen  any  two  sup- 
ports are  independent  of  the  mode  of  distribution  of  the 
moment-area  so  long  as  the  amount  of  that  moment-area  and 
the  position  of  its  centre  of  gravity  remain  unchanged.  This 
relation  is  to  be  of  great  service. 

396.  Re-  Arrangement  of  the  Moment-Area.  —  As  another  illus- 
tration conducing  to  clearness  in  later  constructions,  let 
us  determine  by  still  another  method  the  end-tangents  of 
the  beam  of  §§  394  and  395.  See  Fig.  444.  As  already 
seen,  their  location  is  independent  of  the  arrangement  of 
the  moment-area  between.  Let  us  re-arrange  this  moment- 
area,  viz.,  the  triangle  ACB,  in  the  following  manner  : 

By  drawing  AX  parallel  to  BC,  and  prolonging  BCio  P 
in  the  vertical  through  A,  we  may  consider  the  original 
moment-area  ACB  to  be  compounded  of  the  positive  mom.- 
area  VBXA,  a  parallelogram,  with  its  gravity-vertical 
passing  through  D,  the  middle  of  the  span  ;  of  the  negative 
mom.-area  VCA,  a  triangle  whose  gravity-vertical  passes 


492 


MECHANICS    OF    ENGINEERING. 


through  A  making  ADi=%  AD;  and  of  another  negative 
mom. -area,  the  triangle  ABX,  whose  gravity-vertical  passes 
through  D3  at  one-third  the  span  from  B.  That  is,  the 
(ideal)  positive  load  A  CB  is  the  resultant  of  the  positive 
load  (M.A.)2  and  the  two  negative  loads  (or  upward  pulls) 
(M.A.\  and  (M.A.)3,  and  may  therefore  be  replaced  by  them 
without  affecting  the  location  of  the  end-tangents,  at  A 


and  B,  of  the  elastic  curve  AB.  These  three  moment- 
areas  are  represented  by  arrows,  properly  directed,  in  the 
figure,  but  must  not  be  confused  with  the  actual  loads  on 
the  beam  (of  which,  here,  there  is  but  one,  viz.,  P). 

From  the  given  shapes  and  dimensions,  since  ACB  = 
192,000  (sq.  in.)  (Ibs.),  we  easily  derive  by  geometrical 
principles  : 

(M.A.\=  +576,000  (sq.  in.)  (Ibs.) 
(M.A.\=  -  96,000 
=  -288,000 


Hence,  with  a  pole  distance  El  -4-  n  =  7.11  in.  as  before, 
and  a  "  moment-load-line  "  formed  of  1'2'  =  (M.A.\,  (on 
scale  of  30,000  (sq.  in.  Ibs.)  to  one  inch)  2'3'  =  (M.A.\,  and 
3'4'  —  (M.A.)3,  first  with  a  trial  pole  0",  construct  the  trial 


CONTINUOUS  GIRDER  BY  GRAPHICS.  493 

polygon  A"B",  and  find  n'  in  usual  way  (§  337) ;  then  take 
a  pole  0'  on  the  horizontal  through  n'  and  the  vertical 
TW,  and  draw  the  new  polygon  A'  123  B'.  It  should 
pass  through  B'  on  a  level  with  A',  and  A'\  and  B'3  are  the 
required  end-tangents  (of  the  exagg.  elastic  curve). 

[NOTE — If  B'  were  not  at  the  same  level  as  A',  but 
(say)  0.40  in.  below  it,  B'  should  be  placed  at  ra,  a  distance 

—  =  2  inches  (on  the  paper)  below  its  present  posi- 
tion, (since  the  distance  scale  is  1:20  and  n  =  100,  in  this 
case)  and  the  "  abutment -line  "  of  final  polygon  would  be 
A'm~\. 

Of  course,  this  special  re-arrangement  of  the  moment- 
area  is  quite  superfluous  in  the  present  problem  of  a  dis- 
continuous girder  (not  built  in),  but  considerations  of  this 
kind  will  be  found  indispensable  with  the  successive  spans 
of  a  continuous  girder. 

397.  Positive  and  Negative  Moment- Areas  in  Each  Span  of  a 
Continuous  Girder  (Prismatic).— In  the  foregoing  problem  of  a 
discontinuous  girder  (covering  one  span  only)  not  built  in  at 
the  ends  (otherwise  it  would  be  classed  among  continuous 
girders),  the  moment-curve,  or  equilibrium  polygon  of 
arbitrary  H,  is  easily  found  by  §  389  without  the  aid  of  the 
elastic  curve,  and  the  end  moments  are  both  zero ;  (i.e.,  the 
moment-curve  meets  the  beam  in  the  end- verticals)  but  in 
each  span  of  a  continuous  girder  the  end-moments  are  not 
zero  (necessarily),  and  the  points  in  the  end-verticals  where 
the  moment-curve  must  terminate  (for  an  assumed  H)  can 
not  be  found  without  the  use  of  the  elastic  curve  (or  of 
some  of  its  tangents)  of  the  whole  beam,  dependent,  as  it  is, 
upon  the  loading  on  all  the  spans,  and  the  heights  of  the 
supports. 

Let  Fig.  445  show,  in  general,  any  one  span  of  a  pris- 
matic continuous  girder  (prismatic ;  hence /is  constant), 
between  two  consecutive  supports  A0  and  B0.  Plt  P2,  etc., 
are  the  loads  on  the  span. 

[If  the  displacement  of  An  relatively  to  the  end-tangent 
at  J?0,  and  the  angle  between  the  end-tangents  (of  elastic 


494 


MECHANICS   OF   ENGINEERING. 


curve)  were  known,  the.  moment-curve  or  equilibrium 
polygon  FWG,  (AS  being  the  axis  of  beam)  might  be 
found  by  a  process  similar  to  that  in  §  390,  but  the  elastic 
curves  in  successive  spans  are  so  inter-dependent  that  the 
above  elements  can  not  be  found  directly.] 


I"  fli  ' 

L_LJ-K 


We  now  suppose,  for  the  sake  of  discussion,  that  the 
whole  girder  has  been  investigated  (by  a  process  to  be 
presented)  for  the  given  loads,  spans,  positions  of  supports, 
etc.,  and  then  the  moment-curve  FEWLG  found,  with  the 
corresponding  force-diagram,  for  the  span  in  the  figure 
and  some  arbitrary  H.  The  horizontal  line  AB  represents 
the  axis  of  the  beam  (for  this  purpose  considered  straight 
and  horizontal)  as  an  axis  from  which  to  measure  the 
moment  ordinates. 

_Thus,  the  moment  (of  the  stress-couple)  at  A  is  =  H  X 
AF\  at  B,  Hx  BG  ;  at  E  and  L,  zero  (points  of  inflection). 

Now,  according  to  the  usual  conceptions  of  analytical 
geometry,  we  may  consider  the  portion  EWL,  of  the  mo- 
ment-area, above  AB  as  positive,  and  those  below,  AEF 
and  LB  G,  as  negative ;  but  since  not  one  of  these  three 


CONTINUOUS   GIRDER    BY    GRAPHICS.  495 

areas,  nor  the  position  of  its  gravity-vertical,  is  known  in 
advance,  since  they  are  not  independent  of  the  other 
spans,  a  more  advantageous  re -arrangement  of  the  moment- 
area  may  be  made  thus : 

Join  FG  and  FB,  and  we  may  consider  the  original 
moment-area  replaced  by  the  following  three  component 
areas:  the  positive  moment-area  FE  WLGF  (shaded  by  ver- 
tical lines);  the  negative  triangular  moment-area  AFB\ 
and  the  negative  triangular  moment-area  BFG  (the  nega- 
tive moment-areas  being  shaded  by  horizontal  lines).  (In 
subsequent  paragraphs,  by  positive  and  negative  moment- 
areas  will  be  implied  those  just  mentioned.) 

These  three  moment-areas,  treated  as  loads,  each  applied 
in  its  own  gravity -vertical,  and  considered  in  any  order, 
may  be  used  instead  of  the  real  distributed  moment-area, 
as  far  as  determining,  or  dealing  with,  the  end-tangents  of  the 
elastic-curve  at  A0  and  B0  is  concerned  (§  396),  and  the  fol- 
lowing advantages  will  have  been  gained  : 

(1.)  The  amount  of  the  positive  moment-area,  (M.A.\  in 
Fig.  445,  (depending  on  the  area  lying  between  the  polygon 
FEWG  and  the  abutment-line  FG  of  the  latter),  and  the 
position  of  its  gravity -vertical)  are  independent  of  other  spans, 
and  can  be  easily  found  in  advance,  since  this  moment-area 
and  gravity -vertical  are  the  same  as  if  the  part  of  the  beam 
covering  this  span  were  discontinuous  and  simply  rested  on  the 
supports  A0  and  B(>,  as  in  §  389. 

(2.)  The  gravity-vertical  of  the  left-hand  negative  mo- 
ment-area, (M.A.\,  is  always  one-third  the  span  from  the 
left  end-vertical,  A0A';  that  of  the  other,  (M.A.\,  an  equal 
distance  from  the  right  end-vertical,  B»B'. 

(3.)  The  two  (right  and  left)  negative  moment-areas 
are  triangular,  each  having  the  whole  span  I  for  its  alti- 
tude, and  for  its  base  the  intercept  AF(or  BG)  on  which 
the  end-moment  depends.  Hence,  if  the  amounts  of  thesa 
negative  moment-areas  have  been  found  in  any  span,  we 
may  compute  the  values  which  AF  and  BG  must  have  for 
a  given  H,  and  thus  determine  the  terminal  points  F  and 
G  of  the  moment-curve  of  that  span  (for  that  value  of  H\ 


496  MECHANICS  OF   EXGINEEKIXG. 

For  example,  if  (M.A.\  has  been  found  (by  a  process  not 
yet  given)  to  be  160,000  (sq.  in.)  (Ibs.)  while  AB  —  I  -  160 
in.,  then  the  moment  Mx  which  AF  represents,  is  computed 
from  the  relation 

(M.A.\  =  y2~AB  x  M^ 


If  IT  has  been  chosen  =  100  Ibs.  we  put  HxAF  =  2000 
and  obtain  AF  =  20  inches  of  actual  distance,  so  that  with 
a  scale  of  1:20  for  distances  AF  would  be  one  linear  inch 
of  paper.  (Of  course,  in  computing  BG  the  same  value  of 
H  must  be  used.)  With  H=  100  Ibs.,  then,  and  F  and  G 
as  known  points  of  the  equilibrium  polygon  FEW  G,  it 
is  easily  drawn  by  the  principles  of  §  341. 

We  thus  notice  that  the  amounts  of  the  two  negative 
moment-areas  are  the  only  elements  affected  by  the  con- 
tinuity of  the  girder,  in  this  re-arrangement  of  the  actual 
moment-areas. 

In  the  lower  part  of  Fig.  445  A'  and  B'  represent  the 
extremities  of  the  (exagg.)  elastic  curve,  the  vertical  dis- 
tance B'B"',  of  B'  from  the  horizontal  through  A'  (in  case 
the  two  supports  A0  and  Bn  are  not  at  same  level,  as  we 
here  suppose  for  illustration)  being  laid  off  in  accordance 
with  the  principles  of  the  note  in  §  396. 

NOTE.  —  It  is  now  evident  that  if  the  "false  polygon  "  (as 
it  will  be  called)  A'I23Bf  has  been  obtained  (and  means  for 
doing  this  will  be  given  later)  in  which  the  first  and  last 
segments  are  the  end  tangents  of  the  (exagg.)  elastic  curve, 
and  which  bears  the  same  relation  to  the  three  moment 
areas  just  mentioned,  as  that  illustrated  in  Fig.  444,  we 
may  proceed  further  to  determine  the  amounts  of  (M.A.)l 
and  (M.A.\  as  follows,  by  completing  the  moment-area  dia- 
gram : 

Having  laid  off  the  known  (M.A.\  (or  positive  moment- 
area)  =2'3',  and  ST=EI-±n,  a  line  parallel  to  12  drawn 
through  2',  determines  the  pole  0',  through  which  paral- 


CONTINUOUS  GIKDEB  BY  GRAPHICS. 


497 


leis  to  A'l  and  B'3  will  fix  1'  and  4'  on  the  vertical  SV, 
and  thus  determine  I'2=(M.A.\  and  3'±'=(M.A.}3.  Their 
numerical  values  are  then  computed  in  accordance  with 
the  scale  of  the  moment-area  diagram. 

The  polygon  A'Vl^B'  will  be  called  the  "false  polygon  " 
of  the  span  in  question,  its  end-segments  being  the  end- 
tangents  of  the  elastic  curve. 

398.  Values  of  the  Positive  Moment-Area  in  Special  Cases.— 
For  several  special  cases  these  are  easily  computed,  and 
as  an  illustration,  Fig.  446  shows  a  continuous  girder,  AF 


of  five  spans,  all  six  supports  on  a  level,  and  the  weight  of 
the  beam  neglected.  At  the  extremities  A  and  F,  as  at 
the  other  supports,  the  beam  is  not  built  in,  but  simply 
touches  each  support  in  one  point ;  hence  the  moments  at 
A  and  F  are  zero,  i.e.,  the  moment  curve  must  pass  through 
A  and  Ft  so  that  in  the  first  span  the  left  negative  mo- 
ment-area, and  in  the  fifth  span  the  right  negative  mo- 
ment-area, are  zero.  The  positive  moment-areas  are 
shaded. 

On  the  first  span  is  placed  a  uniformly  distributed  load 
W  over  the  whole  span  I.  .-.  the  positive  moment-area 
for  that  span  is  the  same  as  in  the  case  of  Fig.  235  [see 
(1)  §  397]  and  being  represented  by  a  parabolic  segment 
whose  area  is  two-thirds  that  of  the  circumscribing  rec- 
tangle, its  value  is 


498  MECHANICS   OF   ENGINEERING. 

(M.A.}'l  =  %.rtWl'xl'=l/uWT*        .        .     (1) 

[see  eq.  (2)  §  242],  while  its  gravity- vertical  bisects  the 
span. 

The  only  load  on  the  second  span  is  a  concentrated  one, 
P",  at  distances  h"  and  l^  from  the  extremities  of  the  span; 
hence  the  positive  moment-area  is  triangular  and  has  a 
value 

(M.*X'=#P"W       .       .        .      (2) 

as  in  §  393.  Its  gravity  vertical  may  easily  be  constructed 
as  in  Fig.  443  [see  (1)  §  397]. 

The  third  span  carries  no  load ;  hence  its  positive  mo- 
ment-area is  zero,  and  the  actual  moment-area  is  composed 
solely  of  the  two  triangular  negative  moment-areas  CDH 
and  DHIt  the  moment-curve  consisting  of  the  single 
straight  line  HI. 

The  fourth  span  carries  a  uniform  load  WIV=tulvllvt  and 
.-.  has  a  positive  moment-area 

(M.4.J?=yuW"(F?        ...    (3) 

as  in  eq.  (1),  acting  through  the  middle  of  the  span  (grav- 
ity-vertical). 

Since  the  fifth  and  last  span  carries  no  load,  its  positive 
moment-area  is  zero,  the  moment  curve  being  the  straight 
line  JF,  so  that  the  actual  moment-area  is  composed  of 
the  left-hand  negative  moment-area. 

At  F  it  is  noticeable  that  the  reaction  or  pressure  of  the 
support  must  be  from  above  downward  to  prevent  the 
beam  from  leaving  the  point  F\  i.e.,  the  beam  must  be 
"  latched  down,"  and  the  reaction  is  negative. 

If  the  beam  were  built  in  at  A  (or  F)  the  moment  at 
that  section  would  not  be  zero,  hence  the  left  (or  right) 
neg.  moment-area  would  not  be  zero  in  that  span,  as  in 
our  present  figure.  But  in  such  a  case  the  tangent  of  the 
elastic  curve  would  have  a  knotcn  direction  at  A  (or  F)  and 
the  problem  would  still  be  determinate  as  will  be  seen. 


CONTINUOUS   GIRDER   BY   GRAPHICS.  499 

A' ,  . .  F'  gives  an  approximate  idea  (exaggerated)  of 
the  form  of  the  elastic  curve  of  the  entire  girder.  A  change 
in  the  loading  on  any  span  would  affect  the  form  of  this 
curve  throughout  its  whole  length  as  well  as  of  all  the 
moment  curves. 

NOTE. — It  is  important  to  remark  that  any  two  of  the 
triangular  negative  moment-areas  which  have  a  common 
base  (hence  lying  in  adjacent  spans)  are  proportional  to 
their  altitude  i.e.,  to  the  lengths  of  the  spans  in  which  they 
occur  ;  thus  the  neg.  mom.-areas  (Fig.  446)  GCH&udDCH 
have  a  common  base  CH, 

=r_  } 

r 

(The  notation  explains  itself  ;  see  figure.)  It  also  follows, 
that  the  resultant  of  these  two  neg.  moment-areas  (if  re- 
quired in  any  construction  ;  see  §  400)  acts  in  a  vertical 
which  divides  the  horizontal  distance  between  their  gravity  ver- 
ticals in  the  inverse  ratio  of  the  spans  to  which  they  belong 
[§  21  and  eq.  (4)  above]. 

Hence,  since  this  horizontal  distance  is  /^"~J~K^  their 
resultant  must  act  in  a  vertical  Y'",  whose  distance  from 
the  gravity-vertical  of  GCH  is  %l'",  and  from  that  of 
CHD,  ftl. 

399.  Amount  and  Gravity-Vertical  of  the  Positive  Moment 
Area  of  One  Span  as  Due  to  Any  Loading. — Since  we  can  not 
deal  directly  with  a  continuous  load  by  graphics,  but  must 
subdivide  it  into  a  number  of  detached  loads  sufficiently 
numerous  to  give  a  close  approximation,  let  us  suppose 
that  this  has  already  been  done  if  necessary,  and  that  P1} 
P2,  etc.,  are  the  detached  loads  resting  in  the  span  AB  in 
question  ;  see  Fig.  447. 

Since  [by  (1),  §  397]  the  positive  moment-area  is  the 
same  as  the  total  moment-area  would  be  if  this  portion  of 
the  beam  simply  rested  on  the  extremities  of  the  span,  not 
extending  beyond  them,  we  may  use  the  construction  in  § 
389  for  finding  it,  remembering  that  in  that  paragraph  the 


500 


MECHANICS   OF   ENGIXEEKING. 


oblique  polygon  in  the  lower  part  of  Fig.  439  will  serve 
as  well  as  the  (upper)  one  whose  abutment -line  is  the 
beam  itself,  as  far  as  moments  are  concerned. 

Hence,  Fig.  447,  lay  off  the  load-line  LL',  take  any  pole 
0,  with  any  convenient  pole-distance  H,  and  draw  the 
equilibrium  polygon  FWG.  After  joining  FG,  FWGF 
will  be  the  positive  moment-area  required. 

To  find  its  gravity  vertical,  divide  the  span  AB,  or  FG', 
into  from  ten  to  twenty  equal  parts  (each  —As]  and  draw  a 


FIG.  447. 


vertical  through  the  middle  of  each.  The  lengths  z,,  «2» 
etc.,  on  these  verticals,  intercepted  in  the  moment-area, 
are  proportional  to  the  corresponding  strips  of  moment- 
area,  each  of  width  =  Js,  and  of  an  amount  =  Hzds. 

Form  a  load  line,  SK,  of  the  successive  a's,  and  with 
any  pole  0',  draw  the  equilibrium  polygon  A'E'  (for  the 
z-verticals).  The  intersection,  R,  of  the  extreme  segments, 
is  a  point  in  the  required  gravity -vertical  (§  336). 

The  amount  of  the  moment-area  is  (M,A.\=  2[Hzds] 


CONTINUOUS  GIRDER  BY  GRAPHICS.  501 

For  example,  with  the  span  =Z=120  in.,  subdivided  into 
twelve  equal  ^/s's,  we  have  Js= 10  inches  (of  actual  distance). 
If  H=4  inches  of  paper  and  SK=  J(«)=10.2  inches  of  paper, 
the  force-scale  being  80  Ibs.  to  the  inch,  and  the  distance- 
scale  15  inches  to  the  inch  (1:15),  we  have 

(M.A. )2=  [4  x  80]  x  10  x  [10.2x15]  =489600.  (sq.  in.)(lbs.) 

400.  Construction  of  the  "False-Polygons"  For  All  the  Spans 
of  a  Given  (Prismatic)  Continuous  Girder,  Under  Given  Loading, 
and  With  Given  Heights  of  Supports.— [See  note  in  §  397  for 
meaning  of  "  false-polygon  ".  j  Let  us  suppose  that  the 
given  girder  covers  three  unequal  spans,  Fig.  448,  with 
supports  at  unequal  heights,  and  that  both  extremities  A 
and  D  are  built-in,  or  "  fixed,"  horizontally.  To  clear  the 
ground  for  the  present  construction,  we  suppose  that,  from 
the  given  loading  in  each  span,  the  positive  moment-area 
of  each  span  has  been  obtained  in  numerical  form  [so 
many  (sq.  in.)  (Ibs.)  or  (sq.  in.)  (tons)]  and  its  gravity-ver- 
tical determined  by  §  398  or  §  399  ;  that  the  horizontal 
distances  (i.e.,  the  spans  I',  I",  and  I"  and  the  distance  be- 
tween the  above  gravity-verticals  and  the  supports)  have 
been  laid  off  on  some  convenient  scale ;  that  El  has  been 
computed  from  the  material  and  shape  of  section  of  the 
girder  and  expressed  in  the  same  units  as  the  above  mo- 
ment-areas ;  that  a  convenient  value  for  n  has  been  se- 
lected (since  El-r-n  is  to  be  the  pole  distance  of  all  the 
moment-area-diagrams),  and  that  the  vertical  distances  of 
B,  C,  and  D,  from  the  horizontal  through  A,  have  been 
laid  off  accordingly  (see  note  in  §  396). 

In  the  figure  (448)  verticals  are  drawn  through  the 
points  of  support ;  also  verticals  dividing  each  span  into 
thirds,  since  the  unknown  negative  moment-areas  (sub- 
scripts 1  and  3)  act  in  the  latter  (§  397) ;  and  the  gravity- 
verticals  of  the  known  positive  moment-areas.  The  ver- 
ticals Y',  Y",  V",  and  V"',  are  to  be  constructed  later. 

The  problem  may  now  be  stated  as  follows : 

Given  the  positions  of  the  supports,  the  value  of  El  and  n, 


502 


MECHANICS   OF    ENGINEERING. 


the  fact  that  the  girder  is  fixed  horizontally  at  A  and  D,  tht 
heights  of  supports,  the  location  of  the  gravity-verticals  of  all 
the  positive  and  negative  moment-areas,  and  the  amounts  of  the 
positive  moment-areas  ;  it  is  required  to  Jind  graphically  the 
" false-polygon"  in  each  span. 

The  "  false-polygons,"  viz.:  A 123  B  for  the  first  span  (on 
the  left),  B 123  C for  the  second,  etc.,  are  drawn  in  the  figure 


Pis.  448. 

for  the  purpose  of  discussing  their  properties  at  the  out- 
set. Since  SB  and  BI  are  both  tangent  to  the  elastic  curve 
at  B,  they  form  a  single  straight  line ;  similarly  01  is  but 
the  prolongation  of  3(7.  Also  Al  and  3Z>  must  be  hori- 
zontal since  the  beam  is  built  in  horizontally  at  its  ex- 
tremities A  and  D. 

That  is,  the  three  false  polygons  form  a  continuous 
equilibrium  polygon  A  . . .  D,  in  equilibrium  under  the 
"loads" 

(M.A.y,  (KA.)2't  (M.AX,  etc., 

so  that  we  might  use  a  single  mom. -area-diagram  in  con- 
nection with  it,  but  for  convenience  the  latter  will  be 


CONTINT'OUS    G1KDER  BY   GRAPHICS.  503 

drawn  in  portions,  one  under  each  span,  with  a  pole  dis- 

tance =  il.       • 

n 

Of  this  polygon  A  ...  D,  we  have  the  two  segments  A\ 
and  3Z>  already  drawn,  and  know  that  it  passes  through 
the  points  B  and  C  ;  we  shall  next  determine  by  construc- 
tion other  points  (called  "  fixed  points  ")  p0',  p",  p0",  p'", 
and  PQ'"  in  (the  prolongations  of)  certain  other  segments. 

To  find  the  "fixed  point  "  p^',  where  the  segment  23  in  the 
first  span  cuts  the  vertical  V,  the  gravity  -vertical  of 
(M.A.\f.  The  vertex  p',  or  1,  is  already  known,  being  the 
intersection  of  Al  with  V.  Lay  off  2'  3'  =  (M.A.\r  which  is 
known,  and  take  a  trial  pole  Ot'with  a  pole-distance  El-^n  ; 
join  0t'  2'  and  Ot'  3'.  Draw  120  ||  to  2'0t  to  determine  2o 
on  the  vertical  (M.  A.\'t  then  through  20  a  line  ||  to  0/3'  to 
cut  V  in  PQ.  The  unknown  segment  23  must  cut  V  in 
the  same  point  ;  since  all  positions  of  O/  on  the  vertical 
T'U'  will  result  in  placing^/  in  this  same  point,  and  one 
of  these  positions  must  be  the  real  pole  0'  (unknown). 
[This  is  easily  proved  in  detail  by  two  pairs  of  similar 
triangles]. 

To  determine  the  "fixed-point  "  p",  in  the  prolongation 
of  segment  12  of  second  span.  The  prolongations 
of  the  segments  12  (of  first  span)  and  12  (of  second  span) 
must  meet  in  a  point  k  in  the  (vertical)  line  of  action  of  the 
resultant  of  (M.  A.)J  and  (M.  A.\"  (§  336).  Although  the 
amounts  of  (M.  A.)3'  and  (M.  A.\"  are  unknown,  still  the 
vertical  line  of  action  of  their  resultant  (by  §  398,  Note)  is 

I" 
known  to  be  Y',  a  horizontal  distance  —   to    the    right    of 

o 


.y  ;  hence  Yf  is  easily  drawn.  Therefore,  the  unknown 
triangle  ^31  has  its  three  vertices  on  three  known  verti- 
cals, the  side  £'3  passes  through  the  known  point  p0',  and 
the  side  13  through  the  known  point  B.  Now  by  pro- 
longing PQ  2o  (or  any  line  through  p0')  to  cut  (M.  A.\'  and 
T  in  30  and  &/>  respectively,  joining  30i?  and  prolonging 
this  line  to  cut  (M.  A.\"  in  some  point  10,  and  then  joining 
ko'  10,  we  have  a  triangle  &0'3010  of  which  we  can  make  a 


504  MECHANICS    OF    ENGINEERING. 

statement  precisely  the  same  as  that  just  made  for  k'  3  1 

But  if  two  triangles  [as  F31  and  V  30  10]  have  their 
vertices  on  three  parallel  lines  (or  on  three  lines  which 
meet  in  a  point)  the  three  intersections  of  their  correspond- 
ing sides  must  lie  on  the  same  straight  line  [see  reference 
to  Chauvenet,  §  378  a].  Of  these  intersections  we  have 
two,^90'  and  B  ;  hence  the  third  must  lie  at  the  intersection 
of  the  line  p0f  B  (prolonged)  with  k0'  10,  and  in  this  way  the 
"fixed  point"  p",  a  point  in  &'l  and  .'.  in  the  segment  12 
(of  second  span;  prolonged,  is  found.  Draw  a  vertical 
through  it  and  call  it  V". 

The  fixed  point  p0"  (in  prolongation  of  segment  23  of  sec- 
ond span)  lies  in  the  vertical  V"  and  is  found  from  p"  and 
tne  known  value  of  (M.A.)2"  precisely  &sp0f  was  found  from 
p'.  That  is,  we  lay  off  vertically  2"  3"  =  (M.A.)a",  and  join 
2"  and  3"  to  Ot",  which  is  any  point  at  distance  El  —  n 
to  the  right  of  2"3".  Through  /'  draw  a  line  ||  to  2"  Ot" 
to  cut  (M.  A.\"  in  20,  then  2^"  II  to  Ot"  3"  to  determine 
p"  on  the  vertical  V". 

The  fixed  points  p'"  andp/"  in  the  third  span  lie  in  the 
prolongations  of  the  segments  12  and  23,  respectively,  of 
that  span,  p'"  being  found  from  the  points  p0"  and  C  and 
the  verticals  (M.A.)3",  Y",  and  (M.A.\'"y  in  the  same  manner 
asp"  was  determined  with  similar  data,  while  p0'",  in  the 
same  vertical  V"  as  p'",  depends  on  (M.  A.)2'"  and  its 
gravity  vertical  as  already  illustrated ;  hence  the  d,etail 
need  not  be  given  ;  see  figure. 

In  this  way  for  any  number  of  spans  we  proceed  from 
span  to  span  toward  the  right  and  determine  the  succes- 
sive fixed  points,  until  the  points  p  and  pQ  of  the  last  span 
have  been  constructed,  which  are  p'"  and  p£"  in  our 
present  problem.  Since  p0'"  is  a  point  in  the  segment  23 
(prolonged)  of  the  last  span,  we  have  only  to  join  it  with 
3  in  that  span,  a  point  already  known,  and  the  segment  23 
is  determined.  Joining  the  intersection  2  with  p'"  we 
determine  the  next  segment  21  and  of  coarse  the  vertex  1, 
which  is  then  joined  with  C  and  prolonged  to  intersect 
M.  A.\"  to  fix  the  segment  1(73  and  the  point  3.  Join 


CONTINUOUS  GIKDEll  BY   GRAPHICS.  505 

3  p0",  and  proceed  in  a  similar  manner  toward  the  left, 
until  the  whole  equilibrium  polygon  (or  series  of  "  false 
polygons  ")  is  finally  constructed  ;  the  last  step  being  the 
irrhig  of  2  with  p'. 

401.  Treatment  of  Special  Features  of  the  Last  Problem. —  (1.) 
If  the  beam  is  simply  supported  at  A,  Fig.  448,  instead  of 
built-in,  (M.  A.)'  becomes  zero,  and  the  two  segments  A\ 

d  12  of  that  span  form  a  single  segment  of  unknown 
direction.  Hence,  the  point  A  will  take  the  place  of  p1, 
and  the  vertical  FA  that  of  V. 

(2.)  Similarly,  if  D,  in  the  last  span,  is  a  simple  support 
(beam  not  built  in)  (M.  A.)zr"  becomes  zero,  and  the  seg- 
ments D3  and  32  form  a  single  segment  of  unknown 
direction,  so  that  after  p0"'  has  been  found,  we  join  p" 
and  D  to  determine  the  segment  Z)2  ;  i.e.,  in  this  last  span, 
D  takes  the  place  of  3  of  the  previous  article. 

(3.)  If  the  first  span  carries  no  load  (M.  A.)a'  is  zero,  and 
the  segments  12  and  23  will  form  a  single  segment  23. 
Hence  if  the  beam  is  built  in  at  A,  pj  will  coincide  with 
the  known  point  p'  (i.e.,  1),  while  if  A  is  a  simple  support 
p  and  p'  coincide  with  A,  since,  then,  (M.  A.)L'  is  zero  and 
A  123  is  a  single  segment. 

(4.)  If  the  last  span  is  unloaded  (third  span  in  Fig.  448), 
(M.  A.).2'"  is  zero,  123  becomes  a  single  segment,  and  hence 
PQ'"  will  coincide  with  p'"  ;  so  that  after  p"'  has  been  con- 
structed it  is  to  be  directly  joined  to  3,  if  the  beam  is 
built  in  at  D,  and  will  thus  determine  the  segment  13  ;  or 
to  D,  if  D  is  a  simple  support,  (for  then  (M.  A.)z"'  is  zero 
and  the  three  segments  12,  23,  and  3  D  form  a  single  seg- 
ment.) 

(5.)  If  an  intermediate  span  is  unloaded  (say  the  second 
span,  Fig.  448)  the  positive  mom. -area,  (M.  A.)2",  is  zero, 
123  becomes  a  straight  line,  i.e.  a  single  segment,  and 
therefore  p0"  coincides  with  p"  ;  hence,  when  p"  has  been 
found  we  proceed  as  if  it  were  po". 

402.  To  Find  the  Negative  Mom. -Areas,  the  Mom. -Curves,  Shears, 
and  Reactions  of  the  Supports. — (1.)  Having  constructed  the 


506  MECHANICS   OF   ENGINEERING. 

false  polygons  according  to  the  last  two  articles,  the  nega- 
tive moment-areas  of  each  span  are  then  to  be  found  by  the 
note  in  §  397,  Fig.  445,  and  expressed  in  numerical  form. 

[If  the  positive  inom.-area  of  the  span  is  zero  the  points 
2'  and  3'  will  coincide,  Fig.  445,  and  in  the  case  mentioned 
in  (3),  (or  (4)),  of  §  401,  if  A  (or  D)  were  a  simple  support, 
in  Fig.  448,  the  mom. -area-diagram  of  Fig.  445  would  have 
but  two  rays.] 

(2.)  The  moments  at  the  supports  (or  "end-moments  "  of 
the  respecthe  spans)  depending,  as  they  do,  directly  on 
the  negative  mom. -areas,  can  now  be  computed  as  illustrated 
in  (3)  §  397.  The  fact  that  each  "  end-moment  "  may  be 
obtained  from  two  negative  mom.-areas,  separately,  one  in 
each  adjacent  span  (except,  of  course  at  the  extremities  of 
the  girder)  forms  a  check  on  the  accuracy  of  the  w.ork. 
-  The  two  values  should  agree  within  one  or  two  per  cent. 

(3.)  The  "  moment-curve  "  of  each  span  or  equilibrium 
polygon  formed  from  a  force-diagram  whose  load-line  con- 
sists of  the  actual  loads  on  the  span  laid  off  in  proper 
order,  can  now  be  drawn,  a  convenient  value  for  H  having 
been  selected  (the  same  Hfor  all  the  spans,  that  the  moment- 
carves  of  successive  spans  may  form  a  continuous  line  for 
the  whole  girder);  since  we  may  easily  compute  the  proper 
moment  ordinate  at  each  support  to  represent  the  actual 
moment,  then,  for.  the  H  adopted,  by  (3)  §  397.  The 
moment-curve  of  each  span,  since  we  know  its  two  extreme 
points  and  its  pole-distance  H,  is  then  constructed  by 
§341. 

(4.)  The  shear.  Since  the  last  construction  involves 
drawing  the  special  force-diagram  for  each  span,  with  a 
ray  corresponding  to  each  part  of  the  span  between  two 
consecutive  loads,  the  shear  at  any  section  of  the  beam  is 
easily  found  as  being  the  length  of  the  vertical  projection 
of  the  "  proper  ray,"  interpreted  by  the  force-scale  of  the 
force-diagram,  as  in  §§  389  and  390.  With  the  shears  as 


CONTINUOUS  GIRDER  BY  GRAPHICS. 


507 


FIG.  449. 

librium,  i.e., 


ordinates  a  shear -diagram  may 
now  be  constructed,  if  desired,  for 
each  span.  The  directions  of  the 
shears  should  be  carefully  noted. 
(5.)  Reactions  of  supports.  Let 
us  consider  "  free  "  the  small  por- 
tion of  the  girder,  at  each  point 
of  support,  included  between  two 
sections,  one  close  to  the  support 
on  each  side,  Fig.  449.  Suppose 
it  is  the  support  (7,  and  call  the 
reaction,  or  pressure  at  that  sup- 
port jRc.  Then,  for  vertical  equi- 
36),  we  have 


(6) 


and,  in  general,  the  reaction  at  a  support  equals  the 
(algebraic)  sum  of  the  two  shears,  one  close  to  the  support 
ou  the  right,  the  other  on  the  left.  The  meaning  of  the 
subscripts  is  evident.  In  applying  this  rule,  however,  a 
free  body  like  that  in -Fig.  449  should  always  be  drawn,  or 
conceived ;  for  the  two  shears  are  not  always  in  the  same 
direction  ;  hence  the  phrase  "  algebraic  sum." 

At  a  terminal  support,  as  A  or  F,  Fig.  446,  if  the  beam 
is  not  built  in,  the  reaction  is  simply  equal  to  the  shear 
(since  the  beam  does  not  overhang)  just  as  in  §§  241  and 
243.  Fig.  446  presents  the  peculiarity  that  the  reaction 
of  the  support  F  is  negative,  (as  compared  with  Rc  in  Fig. 
449);  i.e.,  the  support  at  F  must  be  placed  above  the  beam 
to  prevent  its  rising  (this  might  also  be  the  case  at  (7,  or 
Dt  in  Fig.  446,  for  certain  relations  between  the  loads). 

403.  Numerical  Example  of  Preceding  Methods. — As  illustrat- 
ing the  constructions  just  given,  it  is  required  to  investi- 
gate the  case  of  a  rolled  wrought-iron  "  I-beam,"  [a  15- 
inch  heavy  beam  of  the  N.  J.  Steel  and  Iron  Co.,]  extend- 
ing over  four  supports  at  the  same  level,  covering  three 


508  MECHANICS  OF   ENGINEERING. 

spans  of  16,  20,  and  14  feet  respectively,  and  bearing  a 
single  load  in  each  of  the  extreme  spans,  but  a  uniform 
load  over  the  entire  central  span.  As  indicated  thus  : 


20ft.  8ft.          6ft. 


30  tons  40  tons  32  tons 

[This  is  a  practical  case  where  W"  is  the  weight  of  a 
brick  wall,  and  P'  and  P'"  are  loads  transmitted  by  col- 
umns from  the  upper  floors  of  the  building  ;  A  and  D  are 
simple  supports,  and  the  weight  of  the  girder  is  neglected.] 

The  beam  has  a  moment  of  inertia  /  =  707  biquad. 
inches,  and  the  modulus  of  elasticity  of  the  iron  is  E  = 
25,000,000  Ibs.  per  sq.  in.,  =  12,500  tons  per  sq.  in. 

Although  with  a  prismatic  continuous  girder  under 
given  loading,  with  supports  at  the  same  level,  it  may  easily 
be  shown  that  the  moments,  shears,  and  reactions,  to  be 
obtained  graphically,  are  the  same  for  all  values  of  I,  so 
long  as  the  elastic  limit  is  not  surpassed,  still,  on  account 
of  the  necessity  of  its  use  in  other  problems  (supports 
not  on  a  leveD.  we  shall  proceed  as  if  the  value  of  I  were 
essential  in  this  one. 

Selecting  the  inch  and  ton  as  units  for  numerical  work, 
we  have 

El=  12500  x  707.  =  8,837,500  (sq.  in.)  (tons)  while  the 
respective  positive  mom.-areas,  from  eqs.  (2)  and  (3)  of 
§  398,  are  : 

(M.  A.\f  =y2  x  30  X  84  X  108  =  136,080  (sq.  in.)  (tons) 
(M.  A.\"  =  l/a  X  40  x  2402        =  192,000     "      " 
(M.  A.\'"  =  %  X  32  x  96  x  72  -  110,592    "      " 

Adopting  a  scale  of  60,000  (sq.  in.)  (tons)  to  the  linear  inch 
of  paper,  for  mom.  -area  diagrams,  we  have  for  the  above 
mom.-areas  2.27  in.,  3.20  in.,  and  1.84  in.,  respectively,  on 
the  paper,  for  use  in  Fig.  448. 


CONTINUOUS  GIRDER  BY  GRAPHICS.  50!) 

Having  laid  off  the  three  spans  on  a  scale  of  60  inches 
to  the  inch  of  paper,  with  A,  B,  C,  and  D  in  the  same  hori- 
zontal line,  we  find  by  the  construction  of  Fig.  443,  that  the 
gravity -vertical  of  (M.  A.\'  lies  3.6  in.  to  the  left  of  the 
middle  in  the  first  span,  that  of  (M.  A.\'"  4.8  in.  to  the 
right  of  the  middle  of  the  third  span ;  while  that  of 
(M.  A.)2",  of  course,  bisects  the  central  span.  Hence  we 
draw  these  verticals  ;  and  also  those  of  the  unknown  neg- 
ative mom.-areas  through  the  one-third  points ;  remember- 
ing [§  401,  (1)  and  (2)]  that  (M.  A.\'  and  \M.  A.)j"  are 
both  zero  in  this  case. 

Since  El  =  8,837,500  (sq.  in.)  (tons),  it  would  require 
147.29  in.  to  represent  it,  as  pole-distance,  on  a  scale  of 
60,000  (sq.  in.)  (tons)  to  the  inch  ;  hence  let  us  take  n  =  50 
for  the  degree  of  (vertical)  exaggeration  of  the  false  poly- 

FT 

gons,  since  the  corresponding  pole-distance  —  =  2.94  in. 

of  paper  is  a  convenient  length  for  use  with  the  values  of 
(M.  A.\',  (M.  A.\",  etc.,  above  given. 

Following  the  construction  of  Fig.  448,  except  that  p'  is 
at  A,  and  pj"  is  to  be  joined  to  D  (§  401),  (the  student  will 
do  well  to  draft  the  problem  for  himself,  using  the  pre- 
scribed scales,)  and  thus  determining  the  false-polygons, 
we  then  construct  and  compute  the  neg.  mom. -areas 
according  to  §  402  (1),  and  the  note  in  §  397,  obtaining  the 
following  results : 

(M.  A.\',   1.43  in.  of  pap.,  =    85,800  (sq.  in.)  (tons) 
(M.  A.\"t  1.77  "    "      "      =  106,200    "      « 
(M.  A.\"t  1.68  «     "     «      =  100,800    "      " 
(M.  A.\"r,1.17 "     "     "      =    70,200    "     " 

The  remaining  results  are  best  indicated  by  the  aid  of 
Fig.  450.  Following  the  items  of  §  402,  we  find  [(3)  §  397] 
that  the  moment  at  B,  using  (M.  A.\"t  is 

Jf.=  gX 106-200  =  885  inch-tons. 
240  in. 


510  MECHANICS    OF   ENGINEERING. 

[or,  using  (M.  A.\f,  \MB  =  2x85>800=893  in.  tons.] 

X«7d 

Similarly,  M0  =  2x^>8°°  -  840  in.  tons, 
[or,  using  (M.  A.\'",  Mc  =  835.7  in.  tons.] 
Hence,  taking  means,  we  have,  finally, 

JfA=0 ;  JfB=889  in.  tons  ;  Jfc=837.8  ;  MD=0. 

Fig.  450  shows  the  actual  mom.-areas  and  shear-dia» 
grams,  which  are  now  to  be  constructed. 


FIG.  450. 

Selecting  a  value  H  —  20  tons  for  the  pole-distance  of 
the  successive  force-diagrams,  (the  scale  of  distances  being 
5  ft.  (60  in.)  to  the  inch  we  have  [(3)  §  397] 

20  x  ~BG  =  MB  =  889 in. -tons  .-.  BG  =  44.4  in.  of  actual 
distance,  or  0.74  in.  of  paper ;  also  20  x  GK  =  MG  =  837.8 
in.-tons  .'.  GK  =  41.89  in.,  or  0.698  in.  of  paper. 

Having  thus  found  G  and  K,  and  divided  BG  into  ten 
equal  parts,  applying  four  tons  in  the  middle  of  each,  we 
construct  by  §  341  an  equilibrium  polygon  which  shall 
pass  through  0  and  TTand  have  20  tons  as  a  pole-distance. 
(We  take  a  force-scale  of  10  tons  to  the  inch.)  It  will 
form  a  (succession  of  short  tangents  to  a)  parabola,  and  is 
the  moment  curve  for  span  BO.  Similarly,  for  the  single 


CONTINUOUS   GIRDER  BY   GRAPHICS.  511 

loads  P'  and  P'"  in  the  other  two  spans,  we  draw  the 
equilibrium  polygons  AN'G  and  KZ"'D,  for  the  same  H 
as  before,  and  passing  through  A  and  G,  and  K  and  D,  re- 
spectively. 

Scaling  the  moment  -ordinates  NN',  QQ",  and  ZZ"\ 
reducing  to  actual  distance  and  multiplying  by  H,  we  have 
for  these  local  moment  maxima,  M^  =  1008,  M^  =  336,  and 
Mz  =  936,  in.  tons. 

Evidently  the  greatest  moment  is  My  and  .*.  the  stress 
in  the  outer  fibre  at  ^will  be  (§  239) 


=lO.Q  tons  per  sq.  inch  which  is  much 

too  large.  If  we  employ  a  20-inch  heavy  beam,  with  /  = 
1650  biquad.  in.,  the  preceding  moments  will  still  be  the 
same  (supports  att  at  same  level)  and  we  have 

^=1008^0=  6.06  tons  per  sq.  in., 

or  nearly  12,000  Ibs.  per  sq.  in.,  and  is  therefore  safe 
(§183). 

If  three  discontinuous  beams  were  to  be  used,  the  20- 
inch  size  of  beam  (heavy)  would  be  much  too  weak,  in 
each  of  the  three  spans,  as  may  be  easily  shown  ;  hence  the 
economy  of  the  continuous  girder  in  such  a  case  is  readily  per- 
ceived. It  will  be  seen,  however,  that  the  cases  of  conti- 
nuity and  of  discontinuity  do  not  differ  so  much  in  the 
shear-diagrams  as  in  the  moment  curves.  By  scaling  the 
vertical  projection  of  the  proper  rays  in  the  special  force 
diagrams  (as  in  §§  389  and  390)  we  obtain  the  shear  for 
any  section  on  AN,  as  «7AR  (see  Fig.  449  for  notation)  = 
12.3  tons  ;  on  NB,  JBL  =  17.7  tons  ;  from  B  to  C  it  varies 
uniformly  from  «7BR  =  20.3  tons,  through  zero  at  Q,  to  JCIt 
—  19.7  tons  of  opposite  sign.  Also,  for  CZ,  JCR  =  18.6 
tons  ;  and  for  ZD,  J"DL  =  13.4  tons.  Hence,  the  reactions 
of  the  supports  are  as  follows  : 

#A=^AR=12.3  tons  ;  BE=J^+JKK=38.0  tons. 
Jffr=«7CR+JrCL=38.3  tons  ;  jRD=«7DIj=13.4  tons. 


512  MECHANICS    OF    ENGINEERING. 

[In  the  shear-diagram,  the  shear-ordinates  are  laid  off 
bdow  the  axis  when  the  shear  points  down,  the  "free  body  " 
extending  to  the  right  of  the  section  considered,  (as  J"CL  in  Fig. 
449) ;  and  above,  when  the  shear  points  upward  for  the 
same  position  of  the  free  body.] 

If  we  divide  the  max.  shear,  20.3  tons  by  the  area  of  the 
web,  13.75  sq.  in.,  of  the  20-inch  heavy  beam,  (§256),  we 
obtain  1.5  tons  or  3000  Ibs.  per  sq.  in.,  which  is  <  4000 
(§  183).  Notice  the  points  of  inflection,  i',  if',  etc.,  where 
M  is  zero. 

Sufficient  bearing  surface  should  be  provided  at  the 
supports. 

A  swing-bridge  offers  an  interesting  case  of  a  continu- 
ous girder, 

404.  Continuous  Girder  of  Variable  Mom.  of  Inertia.  — If  /  is 
variable  and  I0  denote  the  mom.  of  inertia  of  some  con- 
venient standard  section,  then  we  may  write  7  =  70  -f-  w, 
when  m  denotes  the  number  of  times  I0  contains  /.  In  a 
non-prismatic  beam,  m  is  different  for  different  sections 
but  is  easily  found,  and  will  be  considered  given  at  each 
section. 

In  eq.  (1)  of  §  391,  then,  we  must  put  IQ  -f-  m  in  place  of 
/  and  thus  write 

d2y_[mMdx]  ,-v 

da?"      El,  '    ^  ' 

and  (pursuing  the  same  reasoning  as  there  given)  may 
therefore  say  that  in  a  girder  of  variable  section  if  each 
small  vertical  strip  (Mdx)  of  the  moment-area  be  multiplied  by 
the  value  of  m  proper  to  that  section,  and  these  products  (or  "vir- 
tual mom.-area  strips)  considered  as  loads,  the  elastic  curve  is  an 
equilibrium  polygon  for  those  loads  ivith  a  pole  distance  =  EIQ. 

In  modifying  §  400  for  a  girder  of  variable  section,  then, 
besides  taking  E!Q  -j-  n  as  pole  distance,  proceed  as 
follows : 

Construct  the  positive  mom.-area  for  each  span  accord- 
ing to  §  399 ;  for  each  z  of  Fig.  447,  substitute  mz  (each  z 


CONTIGUOUS    GIRDER    BY    GRAPHICS.  513 

having  in  general  its  own  m),  and  thus  obtain  the  "  virtual 
positive  mom.-area,"  and  its  gravity  vertical. 

Similarly,  there  will  be  an  unknown  "  virtual  neg.  mom.- 
area"  not  triangular,  replacing  each  neg.  mom.-area  of 
§  400.  Though  it  is  not  triangular,  each  of  its  ordi- 
nates  equals  the  corresponding  ordinate  of  the  unknown 
triangular  neg.  mom. -area  multiplied  by  the  proper  m,  and 
its  gravity -vertical  (ichich  is  independent  of  the  amount  of  the 
unknotvn  neg.  mom.-area)  is  found  in  advance  by  the  process 
of  Fig.  447,  using,  for  z's,  a  set  of  ordinates  obtained  thus : 
Draw  any  two  straight  lines  AB  and  FB,  Fig.  445,  (for  a 
left-hand  trial  neg.  mom.-area;  or  FB  and  GFior  a  right- 
hand  one)  meeting  in  the  end-vertical  of  the  span,  divide 
the  span  into  ten  or  twenty  equal  spaces,  draw  a  vertical 
through  the  middle  of  each,  noting  their  intercepts  between 
AB  and  FB.  Add  these  intercepts  and  call  the  sum  S. 
Multiply  each  intercept  by  the  proper  m,  and  with  these 
new  values  as  z's  construct  their  gravity  vertical  as  in  Fig. 
447.  Add  these  new  intercepts,  call  the  sum  Sv,  and  denote 
the  quotient  S  -s-  Sv  by  /?. 

We  substitute  the  three  verticals  mentioned,  therefore, 
for  the  mom.-area  verticals  of  §  400,  and  the  "  virtual  pos. 
mom.-area  "  for  the  pos.  mom.-area,  in  each  span ;  pro- 
ceed in  other  respects  to  construct  the  "false  polygons  " 
according  to  §  400.  Then  the  result  of  applying  the  con- 
struction in  the  note  §  397  will  be  the  "  virtual  neg.  mom.- 
areas,"  each  of  which  is  to  be  multiplied  by  the  proper 
ft  to  obtain  the  corresponding  triangular  neg.  mom.-area, 
with  which  we  then  proceed,  without  further  modifica- 
tions in  the  process,  according  to  (2),  (3),  etc.  of  §  402. 

[The  conception  of  these  "  virtual  mom. -areas  "  is  due 
to  Prof.  Eddy  ;  see  p.  36  of  his  "  ^Researches  in  Graphical 
Statics,"  referred  to  in  the  preface  of  this  work.] 

405.  Remarks. — It  must  be  remembered  that  any  unequal 
settling  of  the  supports  after  the  girder  has  been  put  in 
place,  may  cause  considerable  changes  in  the  values  of  the 
moments,  shears,  etc.,  and  thus  cause  the  actual  stresses  to 
be  quite  different  from  those  computed  without  taking 


514  MECHANICS   OF   ENGINEERING. 

into  account  a  possible  change  in  the  heights  of  the  sup- 
ports. See  §  271. 

For  example,  if  some  of  the  supports  are  of  masonry, 
while  others  are  the  upper  extremities  of  high  iron  or 
steel  columns,  the  fluctuations  of  length  in  the  latter  due 
to  changes  of  temperature  will  produce  results  of  the 
nature  indicated  above. 

If  an  open-work  truss  of  homogeneous  design  from  end 
to  end  (treated  as  a  girder  of  constant  moment  of  inertia, 
whose  value  may  be  formulated  as  in  §  388,)  is  used  as  a 
continuous  girder  under  moving  loads,  it  will  be  subject 
to  "  reversal  of  stress  "  in  some  of  its  upper  and  lower  hor- 
izontal members,  i.e.,  the  latter  must  be  of  a  proper  de- 
sign to  sustain  both  tension  and  compression,  (according 
to  the  position  of  the  moving  loads,)  and  this  may  disturb 
the  assumption  of  homogeneity  of  design.  Still,  if  /  is 
variable,  §  404  can  be  used ;  but  since  the  weight  of  the 
truss  must  be  considered  as  part  of  the  loading,  several 
assumptions  and  approximations  may  be  necessary  before 
3sta  Wishing  saiisfactory  dimensions. 


INDEX  TC  i,lEGH4NIGS  OP 


Abutment-Line  414 

Abutments  of  Arches  430,  435 

Ant i -Re suit ant  402 

Anti-Stres«-Re*ultant  411 

Arches  ^inear  386,  396 

Arches  of  Masonry  421,  437 

Arch-Ribs  438,  483 

/vrch-Ribs, Classification  of  458 

Arch-Rib  of  Three  Hinges  458,  460 

Arch-Rib  of  Hinged  Unas   440,  458,  461 

Arch-Rib  of  Fixed  Ends   439,  459,  465 

Arch  Truss  or  Braced  Arch  478 

Autographie  Testing  Machine  240 
Beams , Rectangular, Compar- 
ative Strength  and  ntiff- 

ness                   272,  273,  277 

Bow's  Notation  407 

Box-Girder  275,  292 

Liraced-Arch  438,  478 

Bridge, Arch  430 

Buckling  of  Web-Flat es  383 
Built  Beams, design ing 

Sections  of  295 

built  Columns  373 

Burr, Prof. . Citations  from  224,  229 

Butt-Joint  226 

Cantilevers             260,  276,  341 

Cantilever, Oblique  353,  356 
C-aten.'.ry  Inverted 

Catenary, Trans formed  395 

Cast  Iron  220,  279 

Cast  Iron, Malleable  224 

Centre  of  Cravity  336 

Chrome  Ste  ^  i224 

Jircle  as  •  -.stic  Curve  262,  343 

Circular  Arc  as  Linear  Arch  391 


Closing  line  414 

Coblenz, Bridge  at  459,  478 

Columns, Long  363 

Compression  of  Short  Blocks  218 

Concurrent  forces  397 
Continuous  Girders, by 

Analysis  32C   332 
Continuous  Girders, by 

Graphics  434   514 

Cover-Plates  226 

Crane  362 

Crank- Shaft, Strength  of  314 

Crushing, Modulus  of  219   424 

Dangerous  Section  262   332 

Deck-Beam  275 

)ef lections, (Flexure  342 

Derivatives, (Elastic  curve)  310 

Diagrams, Strain  209   241 
Displacement  of  Joint  of 

/•.rch  Rib  447 

Dove-Tail  Joint  269 
Eddy, Prof .  , Graphic  methods  , 

(See  Preface)  426 

plastic  Curve  a  Circle  262   343 
Elastic  Curve  an  Equilibrium 

Polygon  484 

Clastic  Curves   245,252,  262    356 
Elastic  Curves, the  Four  x-De- 

rivatives  of  316 

Elasticity-Line  241 

Elasticity, Modulus  of  203,    227 

Elastic  Limit  202 

Elliptical  Beam  340 
Elongation  of  7<Trourht-lron 

Rod  207 

Equilibrium  Polygon  401    45JD 
Equilibrium  Polygon  Through 

Three  Points  418    419 
Exaggeration  of  Vertical 

Dimensions  in  Arch- Ribs  470 

Examples  in  Flexure  280    284 

Examples  in  Shearing  231    232 
Examples  in  Tension  and 

Compression  222    223 
Examples  in  Torsion 
Experiments  of  an  English 

Llroa;  Qommission  314 


3 

'Experiments  of  Hodgkinson   207  369 

Experiments  of  Prof. Lanza  280 

Experiments  on  Building  Stone  424 

Experiments  on  Columns  378 

Ext r ados  421 

Euler's  Formula  for  Columns  364 

Factor  of  Safety  223 

"False  Polygons"            497  501 

Fatigue  of  Metals  224 

"Fixed  Points"  503 

Flexural  Stiffness  250 

Flexure                    244  386 
Flexure  and  Torsion  Con- 

bined  314 
Flexure  Beams  of  Uniform 

Strength  335 

?>. ure, Common  Theory  244 

Flexure, Eccentric  ;>oad      256  301 
Flexure, Mastic  Curves  in 

251,252 

Flexure, Examples  in         280  284 

Flexure, Hydrostatic  Pressure  308 

Flexure, Moving  Loads  298 
Flexure,Kon-Prismatic 

Beams                     332  335 

Flexure  of  Long  Columns  363 
Flexure  of  Prismatic  Beams 

Under  Oblique  Forces      347  362 

Moxura,  Safe  Loads  in      262  284 

Flexure, Safe  Stress  in  279 

^lexure, Shearing  Stress  in   234  295 

Flexure, Special  Problems  in  295  319 

Flexure, Strength  in  249 

plexure,the  Elastic  Forces 

Flexure, the "Moment"  249 

.exure,the  "Shear"  243T 

Flexure, Uniform  Load    258,267  302 

',05,307,324,329  340 

Flow  of  Solids  212 

Force  Diagram  400 

"-lygons 

'orces, Distributed  197 

Friction                   422  423 


General  Properties  of  Mat -rials  204 
Craphic  Treatment  of  Arch  431 
.Gravity,  cent  re  of  336  453 


4 

Gravity, Vertical  453 

Graphical  Statics, Elements  397  420 
Graphical  Statics  of  Vertical 

Forces                 412  420 
Kodgkinson's  Formulae  for 

Columns  369 

-oke's  Law            201,203  207 

ooks, ^trength  of  362 
Horizontal  Straight  Girders 

by  Graphics             479  433 

lorse-fower               239  242 

I-Beam             275,292  295  337 

Inclined  Beam  359 
Internal  Stress, General 

Problem  of  205 

Intrados  421 

I  sot ropes  204 

;iiza,  Experiments  of  Tr^-f.  280 

Lateral  Contraction        211  229 

Lateral  Security  of  Cirders230  298 

Linear  Arches       506,396,417  425 

Live  Loads                298  430 

Load-Line  413 

Locomotive  on  Ar'-h  430 

Locomotice  on  Girder  298 

Malleable  Cast  Iron  224 

Middle  Third  423 

Modulus  of  Elasticity       203  227 

Modulus  of  Resilience  213 

Modulus  of  Rupture (Flexure)  278 

Modulus  of  Tenacity  212 

Moduli  of  Compression  219 

Mohr's  Thoerem  486 

Moment- '.re a  435 

Moment -diagram  263 

Flexure       248,318,  : 

Moment  of  Inertia  by  Graphics  454 

Moment  of  Inertia  of  Kox-Girder  276 
:omcnt  of  Inertia  of  Built  Golumn379 

Moment  of  Inertia  of  Buint  Beam  296 
Moment  of  Inert  is  of  Plane 

Figures                  249  274 

Moment  of  Inertia  of  Truss  478 

Moment  of  Torsion  236 

Mortar  422 

Moving  Loads (Flexure)  298 

Ilaperian'  kajcariikaix  Base     357  337 
Navier's  Principle           422,436 


ireutral  nxis       245,247,  347,  355 

lion-Concurrent  Forces  in  a 

Plane  399 

Normal  Stress  200 
Oblique  Section  of  Rod 

in  Te-.sion  200 

Parabola  as  Linear  Arch  391 

536,344 

Phoenix  Columns  378 

Pier  Reactions  404 

of  Arches  430,435 

"Pin- and- Square" Columns  364 

Polar-Moment  of  Inertia  238 

Polelin  Graphics)  401 

Pole-Distance  416,417 

Practical  Notes  223 
Power, Transmission  of  by 

Shafts  238  318 

Punching  Rivit  Holes  229 

Radius  of  Curvature  250,353 

Radius  of  Gyration  313,376 
Rankine's  Formulae  for 

Columns  372,372 

Rays  of  Force  Diagram  401 

Reaction  4C4 

Reduced  Load- Contour  429 
R&gsdLxfcisra  af  xxakisisx 

Resilience       204,213,237,251,3125 

Reversal  of  Stress  514 

Riveting  fir  Built  Reams  292 

Rivets  and  Riveted  PI  225,292 

Robinson, Prof .Integration  by  357 

Rod  in  Tension  198,200 

Roof  Truss  405 

Rupture  202 

Safe  Limit  of  Stress  202 

Safe  Loads  in  Flexure  262^284 

St. Louis  Bridre  459,467,478 

:et,Parmanent        202,209,208,241 

Shafts  233  239 

Shafts, Non-circular  239 

Shear  Diagram (Flexure-  265 

Shearing  225,232 

Shearing  Distortion  227 

Shear, Distribution  of  in  Flexure  287 

Shearing  Forces  225 
Shearing  Stress    200,201,225,234,284 


.-hear, the  First  x-De- 

rivative  of  Moment (Flexure)  264 
xki&dtxxg 

Slope (inPlexure)  253 

Soffit  421 

Spandrel  421 
Special  Equilibrium 

Polygon            469,424,  440 

Statics, Graphical         397  420 

Stiffening  of  V.'eb-J  lutes  383 

Stone, Strength  of          221  424 

"tress  Diagram  for   rch-Ribs  471 

Stresses  due  for  Sib  Shortening  476 

Strain  Diagrams            209  241 

Strains, t~  o  kinds  only  196 

Stress  '                   197  193 
Stress  and  Strain, ''elation 

Betvcen  201 

Stress  Couple             253  348 

Stress, Nornal  and  Shearing  200 

Strength  of  Materials  195 
Stretching  of  aTrisra  Under 

its   Cwn  r' eight  215 
"Sudden"   Ap^-lic/ition   of  a 

Load                  214  255 
Summation  of  Products  by 

Graphics  451 
Table  for  Flexure  279 
Table  for  Shearing  228 
Tables  for  Tension  and  Com- 
pression 221 
Temperature  Stresses  206,217/222,473 
Tenacity, Moduli  of  212 
Testing  Machine, autograph! s  240 
Theorem  of  Three  Moments  ?32 
Thrust (in  Flexure)  348/350 
Torsion  233,243 
Tors ion, Angle  of 

Tor s ion, Helix  Angle  in  233 

Tor si on, Moment  of  236 

Torsional  Resilience  237 

Torsional  stiffness  236 

Torsional  Strength  235 

Transformed  Catenary  395 
Transmission  of  Tower  by  Shafting 

238,  318 

Trussed  Girders  381 


Page 

'Uniform  Strength  Beams 
Uniform  Strength  Solid  of, 

in  Tension  216 

Voussoir  386   421 

7eb  of  I-3eam  274 

Web  of  I-Beam,bucklinf  of 
"eb  of  I -Beam, Shear  in          290 
W or king- Beam  336   344 

Corking  Strength  202 


This  book  is  DUE  on  the  last  date  stamped  below 


NOV      6  1946 
AUG  1 1  1947 


Form  L-9-15m-7,'31 


TA 
350     Church* 


C47m     Mechanics   of 
— vv2 — engine  erirrg. 


A     001  113692     6 


"550 


AL1FOBS1* 


